Learning Objectives

After going through thin unit, you will be able to understand, appreciate and apply the following concepts:

• The significance and basic details of the alpha particle scattering experiment.

• Rutherford’s model of the atom.

• Bohr’s model; its basic postulates.

• ‘Energy Levels’ and their significance.

• Hydrogen spectrum; its basic details.

• The atomic nucleus.

• Composition and size of the nucleus.

• Atomic masses.

• Isotopes, isobars and isotones.

• The phenomenon of radioactivity.

• The nature and basic properties of alpha, beta and gamma rays.

• Radioactive decay law.

• Mass energy relation.

• Mass defect.

• Binding energy per nucleon and its variation with mass number.

• Nuclear fission.

• Nuclear Fusion.

The Models of the Atom

Introduction: The discovery of the electron and the fact that electron is a common constituent of all matter established the fact that atom has a structure. To assign a model to the atom we have to picture the relative arrangement of positive and negative charges in the atom. In addition a viable model of atom must also explain the neutrality and stability of the atom. Further developments in this field established that under proper conditions, every atom emits a characterstic spectrum. Hence a successful model of the atom should also be able to explain this basic property of the atom.

The Thomson Model

Thomson proposed a first viable model of the atom in 1904. In this model he visualized the positive charge of the atom to be distributed uniformly over the entire atom. The negative electrons were imagined as embedded in this continuous charge. This arrangment is somewhat similar to the arrangement of seeds in a watermelon. Figure shows the arrangement of electrons in atoms with atomic number $z=1$ to $z=4$.

Rutherford’s $\alpha$-Particle Scattering

To experimentally test the viability of Thomson’s hypothesis, Rutherford proposed a historic experiment; the scattering of $\alpha$ particles by thin metal foils.

Rutherford argued that if positive charge is uniformal distributed in atom, then $\alpha$ particles should be able to pass through thin foils with small deviations in their direction of motion. $\alpha$-particles will undergo small angle scattering. Geiger and Marsden then performed the historic experiments to check the basic ideas suggested by Rutherford. Gerger and Marsden used a radioactive element as the source of $\alpha$ particles. A very fine beam of $\alpha$-particles was selected with the help of a lead screen with a small hole in it as shown in the figure. A ZnS screen was placed on the other side of a thin gold foil. Whenever an $\alpha$-particle strikes the screen it produces a flash of light.

Geiger and Marsden discovered that most of a particles passed through gold foil with small deviation. But there were fews $\alpha$-particles that were scattered through very large angles. Very few were even sent backwards! Since $\alpha$-particles are relatively heavy and are travelling with large speeds, it is evident that only very strong force could send them backwards. To explain the results of Geiger and Marsden experiment. Rutherford pictured an atom as follows:

(1) The atom has a small positively charged nucleus. The entire mass and positive charge of the atom is concentrated in the nucleus.

(2) The electrons move around the nucleus.

(3) Since the entire mass of the atom is concentrated in small nucleus most of the atom is empty.

(4) The fact that a very small number of $\alpha$-particles are scattered through large angles futher suggests that the size of atomic nucleus is very small (estimated value: ${10}^{-15}\mathrm{m}$ )

Rutherford Scattering Formula

(1) The distance of closet approach $\left(r_c\right)$ is given by the relation

$${\mathrm{r}}_{\mathrm{c}}={\displaystyle \frac{\mathrm{Ze}(2\mathrm{e})}{4\pi {\epsilon }_0\left(1/2{\mathrm{mv}}^2\right)}}$$

where $1/2{\mathrm{mv}}^2$ is initial K.E of incident $\alpha$-particle. It is obvious from the above relation that the radius of nucleus must be smaller than the calculated value of ${\mathrm{r}}_{\mathrm{c}}$ as an alpha particle cannot touch the periphery of the nucleus on account of strong repulsive force.

(2) Impact Parameter

Impact parameter is defined as the perpendicular distance of the initial velocity vector of the $\alpha$ particle from the central line of the nucleus when the $\alpha$-particle is far away from the nucleus of the atom as shown in Fig. (a).

Fig. (b) shows thr

for particle 1 . The angle of scattering of 1 is more than that of 2 and 3 . This is because particle 1 is coming more close to nucleus $\mathrm{N}$ and therefore experiences more replusive force.

It can be shown that the number or $\alpha$-particle scatrered inrougn an angle $\theta$, is

(1) directly proportional to thickness ( $t$ ) of the gold foil.

(2) directly proportional to atomic number (z) of scattering nuclei.

(3) inversly proportional to kinetic energy (K) of incident $\alpha$-particle.

(4) inversly proportional to ${\sin }^4(\theta /2)$.

Rutherford Atomic Model

According to Rutherford the electrons are revolving around the nucleus in circular orbits (now we postulate eliptical orbits). This is very similar to the planetary model of the universe where planets are revolving round the sun in circular orbits. The attraction between the positively changed nucleus and negative electron provide the centripetal force acting on the electron moving in circular orbit.

For an electron of mass $m$, moving in a circular orbit of radius $r$, with spee $v$;

$${\displaystyle \frac{{\mathrm{mv}}^2}{\mathrm{r}}}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}\left({\displaystyle \frac{\text{ Ze.e }}{{\mathrm{r}}^2}}\right)$$

In the Rutherford model all value of $\mathrm{r}$ and $\mathrm{v}$ are permitted.

Also ${\mathrm{rv}}^2=\left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}\right){\displaystyle \frac{{\mathrm{Ze}}^2}{\mathrm{m}}}=$ const.

$\mathrm{K}=\mathrm{K}$.E of electron in orbit $={\displaystyle \frac{1}{2}}{\mathrm{mv}}^2$

$\mathrm{U}=$ P.E of electron in orbit $=-\left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}\right){\displaystyle \frac{\text{ Ze.e }}{\mathrm{r}}}=-{\mathrm{mv}}^2$

$\mathrm{E}=\mathrm{T}$.E of electron in orbit $=\mathrm{K}+\mathrm{U}=-{\displaystyle \frac{1}{2}}{\mathrm{mv}}^2$

Obviously,

$$\mathrm{K}.\mathrm{E}=|\mathrm{T}.\mathrm{E}|\text{ and }\mathrm{P}.\mathrm{E}=2\mathrm{T}.\mathrm{E}$$

Since $r$ and $v$ are continuous variables, $E$ is also a continuous variable. In other words all values of energy of electron in orbit are possible.

Limitations of the Rutherford Model

(1) Rutherford model faced some serious problems when subjected to the demands of electromagnetic theory. According to classical electromagnetic theory an accelerated charge particle must emit electromagnetic radiation. Now an electron moving in a circular orbit around the nucleus must experience centripetal acceleration and hence in accordance with the demands of electromagnetic radiation will lose energy in the form of electromagnetic radiation. This energy loss will come from the atomic system itself. As a consequence the electron will be continuously pulled towards the nucleus. Hence an electron can not remain in an orbit of radius $r$ but moves along a spiral path as shown in figure.

As the electron goes nearer the nucleus, its speed increases continuously, the centripetal acceleration it experience also becomes more and it must emit more radiations. The net result of all this is that the electron crashes into nucleus after following a spiral path (in about ${10}^{-8}\mathrm{s}$ ). Hence Rutherford model cannot give us a stable atom.

(2) Since electron in orbit losses energy continuously the spectrum of the atom should be a continuous spectrum. However atoms exhibit a characterstic line spectrum. This fact is also ’not in tune’ with the Rutherford model.

However the idea of a planetary atom was hard to discard. Efforts were made to overcome this difficulty which was ultimately successfully explained by Neil Bohr in 1913. However before Bohr’s published his theory a lot of experimental work was already available in the field of spectroscopy. It was established that emission of electromagnetic radiations in the form of spectral lines was the fundamental characterstics of the atoms. With the development in the experimental techniques, the wavelength of emitted spectral lines could be determined very accurately.

Spectrum of Hydrogen Atom

Carefull analysis of the experimental observatinos of the line spectrum emitted by hydrogen showed that it can be described in terms of empirical formulae. The hydrogen spectrum consists of the following series each named after it discoverer.

(1) The Lymann Series

The wave number $\overline{v}={\displaystyle \frac{1}{\lambda }}$ for Lymann series is given by

$$\overline{v}=\mathrm{R}[{\displaystyle \frac{1}{(1{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}];\mathrm{n}=2,3,4,\dots \dots$$

$\mathrm{R}$ is a constant known as Rydberg’s constant. The series lies in the ultra-violet region.

(2) For the Balmer Series

$$\overline{v}=\mathrm{R}[{\displaystyle \frac{1}{(2{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}];\mathrm{n}=3,4,5,\dots \dots$$

The Balmer series lies in the visible part of spectrum. The line for transition from $\mathrm{n}=3,4,5,\dots \dots$. to $\mathrm{n}=2$ are known as ${\mathrm{H}}_{\alpha };{\mathrm{H}}_{\beta }$ and ${\mathrm{H}}_{\gamma }$ lines respectively.

(3) For the Paschen Series

$$\overline{v}=\mathrm{R}[{\displaystyle \frac{1}{(3{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}];\mathrm{n}=4,5,6,\dots \dots$$

The Paschen series lies in the infrared region.

(4) For the Brackett Series

$$\overline{v}=\mathrm{R}[{\displaystyle \frac{1}{(4{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}];\mathrm{n}=5,6,\dots \dots$$

(5) For the Pfund series we have $\overline{v}=\mathrm{R}({\displaystyle \frac{1}{(5{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}});\mathrm{n}=6,7,\dots \dots$

The Brackett and the Pfund series lie in far infrared region.

The Bohr Atom

The postulate of Bohr’s theory are as follows:

(1) Bohr’s First Postulate: An electron revolving around the nucleus is not permitted to move in every possible orbit. It can move only in some “special orbits”. These orbits are known as stationary orbits. An electrons in a stationary orbit cannot lose energy according to the demands of classical electromagnetic theory.

To define these special orbits Bohr further postulated that the angular momentum of an electron in a stationary orbit is always an integral multiple of $\mathrm{h}/2\pi$ where $\mathrm{h}$ is Plancks constant. For an electron of mass m moving with a velocity $v$ in an orbit of radius $r$, we must have

$$\mathrm{mvr}=\mathrm{n}{\displaystyle \frac{\mathrm{h}}{2\pi }}$$

where $\mathrm{n}$ is an integer having permitted values $1,2,3,\dots \dots .\mathrm{n}$ is known as the principal quantum number. Each value of $\mathrm{n}$ corresponds to a stationary orbit.

(2) Bohr’s second postulate is designed to explain the emission of radiations by atoms. This postulate state that whenever an electron jumps from one stationary orbit to another the energy difference between the two is either emitted or absorbed as a quantum of radiation. Expressed mathematically

$${\mathrm{E}}_{{\mathrm{n}}_2}-{\mathrm{E}}_{{\mathrm{n}}_1}=\mathrm{h}\nu$$

where ${\mathrm{E}}_{{\mathrm{n}}_1}$ and ${\mathrm{E}}_{{\mathrm{n}}_2}$ are the energy of electron in the ${\mathrm{n}}_1^{\text{th }}$ and ${\mathrm{n}}_2^{\text{th }}$ orbit and $v$ is the frequency of radiation emitted when electron makes a transition from ${\mathrm{n}}_2^{\text{th }}$ orbit to ${\mathrm{n}}_1^{\text{th }}$ orbit.

Bohr Formule

Let the charge of atomic nucleus be $\mathrm{Ze}$ (for hydrogen $\mathrm{z}=1$ ). When the electron is moving in an orbit of radius ${\mathrm{r}}_{\mathrm{n}}$, the centripetal force maintaining it in this orbit is coulomb electrostatic force

$$\therefore {\displaystyle \frac{{\mathrm{mv}}_{\mathrm{n}}^2}{{\mathrm{r}}_{\mathrm{n}}}}=\mathrm{K}{\displaystyle \frac{\mathrm{Ze}(\mathrm{e})}{{\mathrm{r}}_{\mathrm{n}}^2}};\mathrm{K}={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$$

Multiplyaing both sides of this equation by ${\mathrm{r}}_{\mathrm{n}}^2$.

We have ${\mathrm{m}}^2{\mathrm{v}}_{\mathrm{n}}^2{\mathrm{r}}_{\mathrm{n}}^2=\mathrm{K}\mathrm{m}\mathrm{Z}{\mathrm{e}}^2{\mathrm{r}}_{\mathrm{n}}$

Squarring Bohr’s condition of Ist postutate we have

$${\mathrm{m}}^2{\mathrm{v}}_{\mathrm{n}}^2{\mathrm{r}}_{\mathrm{n}}^2={\mathrm{n}}^2{\displaystyle \frac{{\mathrm{h}}^2}{4{\pi }^2}}$$

Comparing above two equations we have

$\mathrm{km}{\mathrm{Ze}}^2{\mathrm{r}}_{\mathrm{n}}={\mathrm{n}}^2{\displaystyle \frac{{\mathrm{h}}^2}{4{\pi }^2}}$

$\text{ or }{\mathrm{r}}_{\mathrm{n}}={\displaystyle \frac{1}{\mathrm{K}}}{\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4\pi {\mathrm{mZe}}^2}}$ $$ …….(1)

Also we have

$${\mathrm{v}}_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\pi \mathrm{m}{\mathrm{r}}_{\mathrm{n}}}}={\displaystyle \frac{\mathrm{nh}}{2\pi \mathrm{m}}}\times {\displaystyle \frac{\mathrm{k}\times 4{\pi }^2{\mathrm{mZe}}^2}{{\mathrm{n}}^2{\mathrm{h}}^2}}$$

or ${\mathrm{v}}_{\mathrm{n}}=\mathrm{K}{\displaystyle \frac{2\pi {\mathrm{Ze}}^2}{\mathrm{nh}}}$ $$ …….(2)

To calculate the energy of the electron in ${\mathrm{n}}^{\text{th }}$ permitted orbit we notice that this energy will comprise the potential as well kinetic energy of the electron.

Now P.E $=-\underset{\mathrm{r}}{\overset{\mathrm{\infty }}{\int }}\mathrm{k}{\displaystyle \frac{Z{e}^2}{{\mathrm{r}}^2}}\mathrm{dr}=-{\displaystyle \frac{{\mathrm{KZe}}^2}{{\mathrm{r}}_{\mathrm{n}}}}=-{\mathrm{mv}}_{\mathrm{n}}^2[\because {\displaystyle \frac{{\mathrm{KZe}}^2}{{\mathrm{r}}_{\mathrm{n}}}}=-{\mathrm{mv}}_{\mathrm{n}}^2]$

and $K.E={\displaystyle \frac{1}{2}}{\mathrm{mv}}_{\mathrm{n}}^2$

Hence the energy of the electron in the ${\mathrm{n}}^{\text{th }}$ orbit is

$E_n={\displaystyle \frac{1}{2}}m{v}_n^2-m{v}_n^2=-{\displaystyle \frac{1}{2}}m{v}_n^2$

$E=-{\displaystyle \frac{1}{2}}m{\left({\displaystyle \frac{K2\pi Z{e}^2}{nh}}\right)}^2=-K^2{\displaystyle \frac{2{\pi }^{2}m{Z}^2{e}^4}{n^2{h}^2}}$ $$ …….(3)

Using Bohr’s second postulate we can write

$\mathrm{h}\nu ={\mathrm{E}}_2-{\mathrm{E}}_1=-{\mathrm{K}}^2\left({\displaystyle \frac{2{\pi }^2{\mathrm{mZ}}^2{\mathrm{e}}^4}{{\mathrm{h}}^2}}\right)({\displaystyle \frac{1}{{\mathrm{n}}_2^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_1^2}})$

$\nu ={\mathrm{K}}^2{\displaystyle \frac{2{\pi }^2{\mathrm{mZ}}^2{\mathrm{e}}^4}{{\mathrm{h}}^2}}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$

$\text{ or }\overline{v}={\mathrm{K}}^2{\displaystyle \frac{2{\pi }^2{\mathrm{mZ}}^2{\mathrm{e}}^4}{{\mathrm{ch}}^3}}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})={\mathrm{RZ}}^2({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$ $$ …….(4)

where $\mathrm{R}=\mathrm{K}{\displaystyle \frac{{\mathrm{Z}}^2\pi {\mathrm{me}}^4}{{\mathrm{ch}}^3}}$ and is known as Rydberg constant

Putting the value of constants we get the value of $\mathrm{R}$ as $1.099\times {10}^7{\mathrm{m}}^{-1}$

The experimental determined value of $\mathrm{R}$ from spectroscopic measurement is $1.097\times {10}^7$ which remarkably well agrees with the value of R obtained on the basis of Bohr’s posulates.

Energy Level Diagrams

Energy level diagrams are a very useful and convenient way of representing various series of hydrogen spectra.

The energy expression of the electron in ${\mathrm{n}}^{\text{th }}$ Bohr orbit given by Eqn (3). After substituting the values of constants, in terms of electron volts the energy ${\mathrm{E}}_{\mathrm{n}}$ is

$${\mathrm{E}}_{\mathrm{n}}=-{\displaystyle \frac{13.6{\mathrm{z}}^2}{{\mathrm{n}}^2}}\mathrm{e}\mathrm{V}$$

For hydrogen $\mathrm{z}=1$. Fig. shows energy level diagram of hydrogen atom. To construct the energy level diagrams we represents a particular state by a horizontal line. The distance between any two lines on the same scale represent the energy difference between two levels considered. The atom is normally in the unexcited or ground state corresponding to $\mathrm{n}=1$. Whenever absorption of energy take place, it is in an excited state. The atom than makes a transition to lower permitted level with the emission of energy in the form of electromagnetic radiation (a spectral line). The energy level diagram depecits various lines in the Lyman, Balmer, Paschen, Brackett and Pfund series.

Limitations of Bohr atom

Bohr model can also be used for the studies of spectra of hydrogen like elements. Inspite of its tremendous success the Bohr model could not explain the observed fine structure of spectral lines in the hydrogen spectrum.

Example-1:

Which state of a triply ionized beryllium atom has the same orbital radius as that of the ground state of hydrogen.

Show Answer

Solution:

The expression for the radius of hydrogen and hydrogen like atoms is given by

$${\mathrm{r}}_{\mathrm{n}}={\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\pi }^2{\mathrm{kme}}^2}}\text{ and }{\mathrm{r}}_{\mathrm{n}}^{\prime }={\displaystyle \frac{1}{\mathrm{z}}}{\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\pi }^2{\mathrm{kme}}^2}}$$

$\mathrm{z}=1$ for hydrogen and hydrogen like atom for ground state. Also it is given that

$$r_1=r_n^{\prime }$$

$\therefore$ we must have

$$\begin{array}{ll}& {\displaystyle \frac{{\mathrm{h}}^2}{4{\pi }^2{\mathrm{kme}}^2}}={\displaystyle \frac{1}{4}}{\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\pi }^2{\mathrm{kme}}^2}}\\ \text{ or }{\mathrm{n}}^2=4\therefore \mathrm{n}=2\end{array}$$

Thus second orbit of triply ionized beryllium atom has the same radius as the ground state of hydrogen atom.

Example-2:

Find the first excitation potential of ${\mathrm{He}}^{+}$ion.

Show Answer

Solution:

The energy of a hydrogen atom in the nth excited state is given by $-{\displaystyle \frac{13.6}{{\mathrm{n}}^2}}\mathrm{eV}$

Hence the energy of a ionised helium atom would be given by

$$={\mathrm{z}}^2(-{\displaystyle \frac{13.6}{{\mathrm{n}}^2}})=-{\displaystyle \frac{4\times 13.6}{{\mathrm{n}}^2}}=-{\displaystyle \frac{54.4}{{\mathrm{n}}^2}}\mathrm{eV}$$

for first excited state, the transition is from $\mathrm{n}=1$ to $\mathrm{n}=2$ state.

Now for $\mathrm{n}=1$

$${\mathrm{E}}_{{\mathrm{H}}_{\mathrm{c}}^{+}}=-13.6\times 4=-54.4\mathrm{eV}$$

and for $\mathrm{n}=2$

$${\mathrm{E}}_{{\mathrm{H}}_{\mathrm{e}}^{+}}=-{\displaystyle \frac{13.6\times 4}{4}}=-13.6\mathrm{eV}$$

Hence the excitation energy of the ${\mathrm{H}}_{\mathrm{e}}^{+}$ion is $-13.6-(-54.4)=40.8\mathrm{eV}$

If $V_1$ is the first excitation potential for ${\mathrm{H}}_{\mathrm{e}}^{+}$

Then, ${\mathrm{eV}}_1=40.8\mathrm{eV}$

or ${\mathrm{V}}_1={\displaystyle \frac{40.8}{\mathrm{e}}}\mathrm{eV}=40.8\mathrm{Volt}$

Example-3:

The Rydberg constant for hydrogen is $10967700{\mathrm{m}}^{-1}$. Find short and long wavelength limit for Lyman series.

Show Answer

Solution:

For Lyman series we have

$$\overline{\nu }={\displaystyle \frac{1}{\lambda }}=\mathrm{R}({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}})$$

For short wavelength limit we must have $\mathrm{n}=\mathrm{\infty }$

Hence ${\displaystyle \frac{1}{{\lambda }_{\mathrm{S}}}}=\mathrm{R}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{\mathrm{\infty }}})=\mathrm{R}$

$${\lambda }_{\mathrm{s}}={\displaystyle \frac{1}{\mathrm{R}}}={\displaystyle \frac{1}{10967700}}\mathrm{m}=911.6\text{AA}$$

and for long wavelength limit we must have $\mathrm{n}=2$

$\therefore {\displaystyle \frac{1}{{\lambda }_{\mathrm{L}}}}=\mathrm{R}({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}})={\displaystyle \frac{3}{4}}\mathrm{R}$

or

$${\lambda }_{\mathrm{L}}={\displaystyle \frac{4}{3}}\times {\displaystyle \frac{1}{\mathrm{R}}}={\displaystyle \frac{4}{3}}{\displaystyle \frac{1}{10967700}}\mathrm{m}=1215\text{AA}$$

Example-4:

Show that ${\displaystyle \frac{\mathrm{n}(\mathrm{n}-1)}{2}}$ different wavelength are observed in the hydrogen spectra if the atoms are excited to states with principal quantum numbers.

Show Answer

Solution:

The Possible states when transition occur from ${\mathrm{n}}^{\text{th }}$ state are $(\mathrm{n}-1{)}^{\text{th }}\dots ..{.2}^{\text{nd }}$ state or $1^{\text{st }}$ state, so there are $(\mathrm{n}-1)$ possible transitions starting from ${\mathrm{n}}^{\text{th }}$ state.

Further the atoms reaching the $(\mathrm{n}-1{)}^{\text{th }}$ state will make $(\mathrm{n}-2)$ different transitions. We can extend this reasoning further to other lower states.

Hence the possible number of transitions

$$\begin{array}{rl}& (\mathrm{n}-1)+(\mathrm{n}-2)+(\mathrm{n}-3)+\dots \dots \dots \dots +2+1\\ & ={\displaystyle \frac{\mathrm{n}(\mathrm{n}-1)}{2}}\end{array}$$

Radioactivity

Radioactivity is the phenomenon in which we have spontaneous disintegration of the nucleus of one or more atoms with the emission of highly pentrating radiations. The phenomenon of radioactivity is confined entirely to heavier elements in the periodic table having atomic number ranging from 83 to 102 .

The phenomenon of radioactivity was accidently discovered by Becquerel. The phenomenon is independent of the external agents like temperature, pressure, electric or magnetic field and the ‘physical’ or chemical state of the specimen. It was M. Curie who gave the name raioactive to all substances capable of emitting these rays. A simple experiment by Rutherford established that the radioactive radiations consists of three components known as $\alpha$, particles $\beta$ rays and $\gamma$ rays. $\alpha$ particles are nothing but doubley ionized helium atoms, $\beta$ rays are electrons moving with velocities compariable to the velocity of light and $\gamma$ rays are electromagnetic radiations of very short wavelengths. We now discuss in some details the properties of each of these three components.

Alpha Particles

1. $\alpha$ particle produce flourescene in zinc sulphide over a thin glass disc. A significant point was that the floursent disc does not show emission of light uniformly but only a series of scintillations, showing that $\alpha$ rays consists of particles which can be counted.

2. $\alpha$ particles are doubely ionized helium atoms. The speed of $\alpha$ particles is about $c/100$; $\mathrm{c}$ (= speed of light).

3. $\alpha$ particles ionize heavily the gases through which they travel.

4. They are easily absorbed by matter.

5. They are scattered while passing through metal sheets.

6. They can produce artificial disintegration of an atom.

7. $\alpha$ particles are chaaracterized by the fact that after they have proceeded a certain distance in air, they are no longer able to ionize the air. This is expressed by saying that they have a well defined range. The range, $\mathrm{R}$ is related with the velocity ’ $\mathrm{v}$ ’ of the $\alpha$ particle by the relation

$${\mathrm{v}}^3=\mathrm{kR}\text{ (empirical relation })$$

The range represents the distance $\alpha$ particles travel before their velocities are reduces by numerous collisions to such a low value that they are no longer capable of ionizing the atoms of the gas concerned. The range ’ $R$ ’ of $\alpha$ particles can be very easily and accurately determined experimentally.

Beta Particle

1. $\beta$ particles are negatively charged particles. There $\mathrm{e}/\mathrm{m}$ values determinations show that they are very fast moving electrons. Speed of $\beta$-particles is less thatn $\mathrm{c}$ but comparable to $\mathrm{c}$.

2. $\beta$ particles are emitted with a range of velocities whose maximum is a characterstics of the isotope from which they emanate.

3. They have less ionizing power as compared to $\alpha$ particles.

4. They originate from the nucleus of disintegrating atom.

5. They do not show any definite range.

6. Their penetrating power is greater than that of $\alpha$ particles

7. $\beta$ particles coming out of a radioactive substance strike against the outer electrons of atom to produce $\gamma$ rays.

8. Nucleus showing $\beta$ emission decay spontaneously by emitting either an electron or a positron. For example

$\stackrel{32}{\underset{15}{\text{P}}}\to \stackrel{32}{\underset{15}{\text{S}}}+\stackrel{0}{\underset{-1}{\text{e}}}+\bar{\nu }$

and $\stackrel{22}{\underset{15}{{\text{N}}_{\mathrm{a}}}}\to \stackrel{22}{\underset{10}{{\text{N}}_{\mathrm{e}}}}+\stackrel{6}{\underset{+1}{\text{e}}}+\nu$

$\overline{\nu }$ and $\mathrm{v}$ represent antineutrions and neutrions. It is believed that they are emitted by the following conversion.

$$\mathrm{n}\to \mathrm{p}+\stackrel{0}{\underset{-1}{\text{e}}}+\bar{\nu }\text{ and }\mathrm{p}\to \mathrm{n}+\stackrel{0}{\underset{+1}{\text{e}}}+\nu$$

Gamma Rays

1. Gamma rays are electromagnetic waves of very short wavelength. Speed of $\gamma$-rays equals $c$.

2. Gamma rays are often emitted in conjunction with $\beta$ particles.

3. They produced flourescene and affect photographic plates.

4. They are reflected from the surface of a crystal. This property of $\gamma$ rays is used to measure their wavelength.

5. They ionize gases very slightly.

6. $\gamma$ rays have greater pentrating power than $\alpha$ or $\beta$ rays.

According to the development in the field of nuclear models it is now known that there are energy levels in the nucleus just like atoms (of the order of $\mu \mathrm{eV}$ ). When the nucleus is in an excited satte it can make a transition to a lower state by the emission of electromagnetic radiations.

Most radioactive nucleus after an alpha or beta emission leave the resultant nucleus in an excited state. This nucleus known as daughter nucleus reaches the ground state by a single or sometimes successive transition by emitting one or more gamma rays.

Laws of Radioactive Disintegration

The laws of radioactivity are brodally divided into two parts

(1) the qualitative law

and (2) the quantitative law

1. The Rutherford-Soddy’s Law

is the qualitative law. It tells the nature of daughter nucleus obtained in a ratioactive transformation.

(a) A particular radioactive nuecleus can emit either an $\alpha$ or a $\beta$ particle but both can never be emitted simuitaneously.

(b) When a radioactive nucleus ${}_Z^A\mathrm{X}$ emits an $\alpha$-particle its mass number decreases by 4 and atomic number by 2 ; i.e.

$${}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}\to {}_{\mathrm{Z}-2}^{\mathrm{A}-4}\mathrm{Y}+{}_2^4\mathrm{He}$$

When a radioactive nucleus emits a $\beta$-particle, the mass number remains same and atomic number increases or decreases by one depending on whether the $\beta$-particle emitted is an electron or a positron.

$$\begin{array}{rl}& {}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}\to {}_{\mathrm{Z}+1}^{\mathrm{A}}\mathrm{Y}+{}_{-1}^0\mathrm{e}(\beta -\text{ particle })\\ \text{ or }& {}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}\to {}_{\mathrm{Z}-1}^{\mathrm{A}}\mathrm{Y}+{}_{+1}^0\mathrm{e}(\text{ Positron })\end{array}$$

2. The Exponential Law

This is quantative law. It relates the number of radioactive atoms initially with the number of atoms present at $\mathrm{t}=\mathrm{t}$.

Hence it becomes obvions that radioactive disintegration are governed by the laws of probability. The phenomenon can only be described statistically.

Let ${\mathrm{N}}_0$ be the number of radioactive atoms at the instant $\mathrm{t}=0$ and let $\mathrm{N}$ be their number at a latter instant $t$. Let $\lambda$ represent the probability that one of these atoms will disintegrete in a unit time. $\lambda$ is known as disintegration constant. Hence the probable number of disintegration per unit time is $\lambda \mathrm{N}$. Thus we have

$$-{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}=\lambda \mathrm{N}\left({\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}\text{ is representing the rate of disintegration }\right)$$

Hence ${\displaystyle \frac{\mathrm{dN}}{\mathrm{N}}}=-\lambda \mathrm{t}$

Integrating we have

$$\begin{array}{rl}& \underset{{\mathrm{N}}_0}{\overset{\mathrm{N}}{\int }}{\displaystyle \frac{\mathrm{dN}}{\mathrm{N}}}=-\lambda \underset{0}{\overset{\mathrm{t}}{\int }}\mathrm{dt}\\ & \text{ or }\ell \mathrm{n}\left({\displaystyle \frac{\mathrm{N}}{{\mathrm{N}}_0}}\right)=-\lambda \mathrm{t}\\ & \therefore \mathrm{N}={\mathrm{N}}_0-{\mathrm{e}}^{\lambda \mathrm{t}}\end{array}$$

which means that radioactive disintegrations are governed figure. $\lambda$ the disintegration constant, is a characterstics of a particular radioactive material.

The Half Life and the Mean Life

It is clear that a radioactive element will lose its activity after a infinite time and all radioactive element are identical in this respect. To distinguish one radioactive element from other as far as its activity is concerned we use the concept of half life. Half life $T$ of a radioactive element is the time in which the initial number of radioactive atoms is reduced to half, i.e. $\mathrm{N}={\displaystyle \frac{{\mathrm{N}}_0}{2}};\mathrm{t}=\mathrm{T}$. Therefore

$${\displaystyle \frac{{\mathrm{N}}_0}{2}}={\mathrm{N}}_0{\mathrm{e}}^{-\lambda \mathrm{T}}\text{ or }\mathrm{T}={\displaystyle \frac{0.693}{\lambda }}$$

We define the concept of mean life to represent the above mentioned facts from a statistical point of view. We have already seen that it it only after an infinite time that the radioactivity of an element disappear completely. Hence out of disintergrating atoms some will disintegrate almost immediately and some other will disintegrate after an infinite time.

Thus actual life of a disintagrating atom can vary from 0 to $\mathrm{\infty }$. We thus introduce the numerical average of the ages of the atoms as the number of atoms decreases from ${\mathrm{N}}_0$ to 0 . Mean life is then the ratio of the combined ages of all the atoms and the total number of the atoms in the disintegratinng atom. Refering to figure we can write, $\tau$, the mean life of a radioactive element as

$$\tau ={\displaystyle \frac{1}{{\mathrm{N}}_0}}\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{t}\mathrm{dN}\text{ or }\tau ={\displaystyle \frac{1}{{\mathrm{N}}_0}}\underset{0}{\overset{\mathrm{\infty }}{\int }}\lambda {\mathrm{N}}_0{\mathrm{e}}^{\lambda \mathrm{t}}\mathrm{dt}$$

on inegrating above expression we have $\tau ={\displaystyle \frac{1}{\lambda }}$

We thus have $\mathrm{T}=0.693\tau$

The curie is the unit of radioactive disintegration. We have

1 curie $=3.7\times {10}^{10}$ disintegration per second

Example-5:

A radioactive sample has $6\times {10}^{18}$ active nuclei at a certain instant of time. How many of these nuclei will still be in the same active state after four half lives?

Show Answer

Solution:

By the definction of half life we know that it represents the time after which the number of original atoms is reduced to half. Hence after one half life the number remaining is $1/2$ of the original number. Thus after two half lives, the active remaining number will be $(1/2)\times 1/2$ of the original number. Proceeding with the same argument the active remaining number of atoms after four half lives will be $1/2\times 1/2\times 1/2\times 1/2$ of the original number i.e.

$$6.0\times {10}^{18}\times {\displaystyle \frac{1}{16}}=0.375\times {10}^{18}$$

Example-6:

The half life of is ${}^{198}\mathrm{Au}$ is 2.7 days. Calculate the activity of $1.00\mathrm{mg}$ of ${}^{198}\mathrm{Au}$. (Atomic weight of $\mathrm{Au}={198}^{\mathrm{mol}-1}$ )

Show Answer

Solution:

The number of atoms in $1.00\mathrm{mg}$ of ${}^{198}\mathrm{Au}$ is given by

$$\mathrm{N}=6\times {10}^{23}{\displaystyle \frac{1.00\mathrm{mg}}{198\mathrm{g}}}=3.03\times {10}^{18}$$

Since half life of ${}^{198}\mathrm{Au}$ is 2.7 days, we have

$$\lambda ={\displaystyle \frac{0.693}{2.7\times 24\times 3600}}=2.9\times {10}^{-6}{\mathrm{s}}^{-1}$$

Now we know that $\left|{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}\right|=\lambda \mathrm{N}$

Hence activity of $1.00\mathrm{mg}$ of ${}^{198}\mathrm{Au}$ will be given by

$$(2.9\times {10}^{-6})\times (3.03\times {10}^{18})\text{ disintegration per second }$$

$$={\displaystyle \frac{2.9\times {10}^{-6}\times 3.03\times {10}^{18}}{3.7\times {10}^{10}}}\text{ curie }=240\text{ curie }$$

Example-7:

One gram of radon has an activity of one curie. What activity of radon will accumulate from one milligram of pure radon in $\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{2}\mathbf{5}$ day which is the half life of radon.

Show Answer

Solution:

No. of atoms in $1\mathrm{mg}$ of pure radion $={\displaystyle \frac{6.023\times {10}^{23}}{222}}\times {10}^{-3}$

Half life of radon $=3.825$ days

$$\lambda ={\displaystyle \frac{0.693}{3.825\times 24\times 60\times 60}}{\mathrm{s}}^{-1}$$

Activity of $1\mathrm{mg}$ of radon is ${\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}=\lambda \mathrm{N}$

$$\begin{array}{rl}{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}=& {\displaystyle \frac{0.693\times 6.02\times {10}^{23}}{3.825\times 24\times 60\times 60\times 222}}\text{ disintegration }\\ & ={\displaystyle \frac{0.693\times 6.02\times {10}^{23}}{3.825\times 24\times 60\times 60\times 222\times 3.7\times {10}^{10}}}\\ & =153.7\text{ curie }\end{array}$$

Example-8:

The activity of a radioactive sample is $\mathbf{4}\mathbf{7}\mathbf{5}\mathbf{0}$ per minute. Five minutes later the activity is 2700 per minute. Find the half life of radioactive substance.

Show Answer

Solution:

Activity of the sample in the begining $=4750$

Activity of the samle after 5 minutes $=2700$

$\therefore 2700=4750{\mathrm{e}}^{5\lambda }$

or ${\mathrm{e}}^{5\lambda }={\displaystyle \frac{4750}{2700}}=1.76$

$\lambda ={\displaystyle \frac{0.5654}{5}}=0.1131{\mathrm{m}}^{-1}$

$\mathrm{T}={\displaystyle \frac{0.693}{\lambda }}={\displaystyle \frac{0.6931}{0.1131}}=6.1$ minute

Example-9:

Calculate the weight of one curie of RaB from its half life of 26.8 minute.

Show Answer

Solution:

Activity of $\mathrm{RaB}={\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}=1$ curie $=3.7\times {10}^{10}$

Half life of $\mathrm{RaB}=26.8$ minute

$$\lambda ={\displaystyle \frac{0.693}{26.8\times 60}}{\mathrm{s}}^{-1}$$

Let $\mathrm{N}$ be the number of atoms of $\mathrm{RaB}$ having an activity of 1 curie, then

$$\begin{array}{c}{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}=\lambda \mathrm{N}\\ \mathrm{N}={\displaystyle \frac{3.7\times {10}^{10}\times 26.8\times 60}{0.693}}\end{array}$$

We know that $6.02\times {10}^{23}$ atom $=1$ gram atom $=214\mathrm{g}$

Weight of RaB having an activity of 1 curie

$$\mathrm{N}={\displaystyle \frac{214\times 3.7\times {10}^{10}\times 26.8\times 60}{6.02\times {10}^{23}\times 0.693}}\mathrm{g}=30.52\times {10}^{-7}\mu \mathrm{g}$$

Example-10:

When the nucleus of copper $(A=65,Z=29)$ is bombarded with neutrons, $\gamma$ rays are emitted and the resultant nucleus is radioactive emitting $\beta$ rays. If the disintegration constant $\lambda$ is 0.00231 ${\mathbf{s}}^{-1}$, calculate the half life and average life of the radioactive product.

Show Answer

Solution:

$$\text{ Half life }\mathrm{T}={\displaystyle \frac{0.693}{\lambda }}={\displaystyle \frac{0.693}{0.00231}}=300\mathrm{s}[\lambda =0.00231{\bar{\mathrm{s}}}^1]$$

Mean life $\zeta ={\displaystyle \frac{1}{\lambda }}={\displaystyle \frac{1}{0.00231}}=432.95\mathrm{s}$

Atomic Nucleus

We know nucleus of an atom has the entire mass and positive charge of atom. The size of nucleus is about ${10}^{-15}\mathrm{m}$. Careful experimental measurements showed that mass of different nuclei is nearly an integral multiple of mass of a proton. We also now that $\beta$-particle emitted by a radioactive nucleus originate from the nuclei. Prout proposed “proton electron model” of atomic nucleus.

The Proton-electron Model

According to Prout a nucleus ${}_Z^{A}X$; $(A=$ mass number; $Z=$ atomic number) consists of $A$ protons and (A-Z) electrons, i.e.

$${}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}=\mathrm{A}\left({}_1^1\mathrm{H}\right)+(\mathrm{A}-\mathrm{Z}){}_{-1}^0\mathrm{e}$$

The hypothesis had to be rejected because the theoretical and experimental values of (1) nuclear spin and (2) nuclear magnetic moment are not in accord with theoretical calculated values and experimental values. Also uncertaintity principle is violated if electrons exist inside nucleus.

Nuclear Reactions

With the development of high energy particle accelerators, proton and $\alpha$-particle having very high energy were used as bullets to bombared a nucleus. In an experiment Rutherford bombard ${}_7^{14}\mathrm{N}$ with a proton $\left({}_1^1\mathrm{H}\right)$ and obtained an isotopes of oxygen. The reaction is represented as

$${}_7^{14}\mathrm{N}+{}_1^1\mathrm{H}\to {}_8^{15}\mathrm{O}$$

This is known as a nuclear-reaction. This is also known as transmutation of elements. We can now represent a general nuclear reaction as:

${}_{Z_1}^{A_1}\mathrm{X}+{}_{Z_2}^{A_2}\mathrm{Y}\to {}_{Z_3}^{A_3}\mathrm{C}+{}_{Z_4}^{A_4}\mathrm{D}+\mathrm{Q}$

In any nuclear reaction mass number and atomic number are conserved i.e.

${\mathrm{A}}_1+{\mathrm{A}}_2={\mathrm{A}}_3+{\mathrm{A}}_4$

and ${\mathrm{Z}}_1+{\mathrm{Z}}_2={\mathrm{Z}}_3+{\mathrm{Z}}_4$

$\mathbf{Q}$ is energy released or absorbed in the reaction. It is known as $\mathbf{Q}$-factor of reaction. In the above reaction $\mathrm{X}$ and $\mathrm{Y}$ are reactants and $\mathrm{C}$ and $\mathrm{D}$ are the products. Let $\mathrm{M}$ and ${\mathrm{M}}^{\prime }$ be sum of masses of reactants and products respectively. If ${\mathrm{M}}^{\prime }<\mathrm{M}$; there is release of energy in reaction and $\mathrm{Q}$ is a positive number, given by $\mathrm{Q}=(\mathrm{M}-{\mathrm{M}}^{\prime })\times {\mathrm{c}}^2$

If however ${\mathrm{M}}^{\prime }>\mathrm{M}$; energy is absorbed in the reaction $\mathrm{Q}$ is a negative number.

Atomic Mass Unit

One atomic mass unit; $1\mathrm{u}$; is $(1/12)$ of mass of an atom of ${}_6^{12}\mathrm{C}.1\mathrm{u}\simeq 1.667\times {10}^{-27}\mathrm{Kg}$.

In accordance with Eintesin’s mass-energy equivalence relation $\mathrm{E}=(\mathrm{\Delta }\mathrm{m})\times {\mathrm{c}}^2$

Energy equivalent to 1 a.m.u of mass $=1.667\times {10}^{-27}\times {(3\times {10}^8)}^2\mathrm{J}$

$$={\displaystyle \frac{1.667\times {10}^{-27}\times {(3\times {10}^8)}^2}{1.6\times {10}^{-13}}}\mathrm{MeV}\simeq 931\mathrm{MeV}$$

The Discovery of Neutron

Bathe and Becker carried out a nuclear reaction by bombarding Brylium $\left({}_4^9\mathrm{Be}\right)$ by $\alpha$-particles $\left({}_2^4\mathrm{He}\right)$. Highly penetrating radiations were emitted. Assuming these radiations to be $\gamma$-rays; Bathe and Becker represented nuclear reaction as

$${}_4^9\mathrm{Be}+{}_2^4\mathrm{He}\to {}_6^{13}\mathrm{C}+\gamma -\text{ rays }$$

Applying the “mass-energy” conservation principle the theoretical energy of emitted $\gamma$-rays works out to be $\sim 5\mathrm{MeV}$. However contradictions were obtained when these radiations were used to bombard a hydrogeneous material and the energy of emitted protons was measured experimentally. Apparently law of conservation of mass-energy was violated! Chodwick overcame all contradictions by predicting that highly penteration radiations observed are not $\gamma$-rays. He proposed to existence of a new particle having almost same mass as proton but no charge. The particle was named as Neutron $\left({}_0^1\mathrm{n}\right)$. According to Chadwick; the correct reaction is

$${}_4^9\mathrm{Be}+{}_2^4\mathrm{He}\to {}_6^{12}\mathrm{C}+{}_0^1\mathrm{n}$$

The Proton-Neutron Hypothesis

Immediately after the existence of neutron was experimentally confirmed physicst proposed the protonmeutron hypothesis of atomic nucleus. Protons and neutrons taken together are referred to as nucleons. According to this hypothesis a nucleus ${}_Z^{\mathrm{A}}\mathrm{X}$ is made up of z protons and (A-Z) neutrons

$${}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}=\mathrm{Z}\left({}_1^1\mathrm{H}\right)+(\mathrm{A}-\mathrm{Z}){}_0^1\mathrm{n}$$

The theoretical and experimental values of(1) nuclear spin and (2) nuclear magnetic moment are in agreement with theortical calculated and experimental values. Also the uncertaintity principle is not violated. The hypothesis also easily explains existance of isotopes; isobars etc. It is the presently accepted model of atomic nucleus.

Size of the Nucleus

We have already discussed the “parameter” the distance of closeset approach in Rutherford scattering formula. The value of ${\mathrm{r}}_0$ varies from 1.2 to 1.5 fermi depending upon the nature of bombarding particle. In Rutherford model of the atom we visulize that the most of the space in atom is empty and the size of the nucleous is small (estimated to be about ${10}^{-15}\mathrm{m}$ ).

We assume nucleus to be spherical in shape and its size increases with the increase of number of nucleons. Thus the volume of a nucleus increases with its mass number A. Being a sphere of radius A, the volume of the nucleus viz. $4/3\pi r^3$, so that radius is proportional to the cube root of its volume, which is proportional to its mass number A. It is found that very approximately

$$r=r_0{A}^{1/3}$$

where $r_0$ is the distance of nearest approach to the nucleus and is sometimes referred as the nuclear unit radius.

Example-11:

Find the maximum energy that a $\beta$-particle can have in the following reaction.

${\mathbf{L}\mathbf{u}}^{176}\to \stackrel{176}{{\text{H}}_{\mathrm{f}}}+\stackrel{0}{\underset{-1}{\text{e}}}+\bar{v}$

Acomic mass of ${\mathrm{Lu}}^{176}$ is $175.942694\mathrm{u}$ and that of ${\mathrm{H}}_{\mathrm{f}}$ is $\mathbf{1}\mathbf{7}\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{4}\mathbf{1}\mathbf{4}\mathbf{2}\mathbf{0}\mathrm{u}$.

Show Answer

Solution:

Mass defect $\mathrm{\Delta }\mathrm{m}=(175.942694-175.941420)\mathrm{amu}$

Kinetic energy available in the given reaction is

$$\begin{array}{rl}& (175.942694-175.941420)\times 931\mathrm{MeV}\\ & =1.182\mathrm{MeV}\end{array}$$

This energy available is to be shared between emitted $\beta$-particle and $\overline{v}$. However $\overline{v}$ will not practically have any energy and practically whole energy i.e. $1.182\mathrm{MeV}$ goes to $\beta$-particle.

Example-12:

A nucleus with $\mathrm{A}=235$ splits into two new nuclei whose mass numbers are in the ratio of 2:1. Find the radii of the new nuclei.

Show Answer

Solution:

We know that

$r=r_0\mathrm{A}1/3$ where ${\mathrm{r}}_0=1.4$ fermi

Let $r_1$ and $r_2$ denote the radii of new nuclei formed and $A_1$ and $A_2$ denote their mass number.

$$\begin{array}{c}{\mathrm{A}}_1={\displaystyle \frac{1}{3}}(235)\\ \therefore {\mathrm{r}}_1=(1.4\mathrm{fm}){\left({\displaystyle \frac{235}{3}}\right)}^{1/3}=5.99\mathrm{fm}\\ \text{ and }{\mathrm{A}}_2={\displaystyle \frac{2}{3}}(235)\\ \therefore {\mathrm{r}}_2=(1.4\mathrm{fm}){\left({\displaystyle \frac{2}{3\times 235}}\right)}^{1/3}=7.55\mathrm{fm}\end{array}$$

Example-13:

There is a stream of neutrons with a kinetic energy of $0.0327\mathrm{eV}$. If the half life of neutron is $700\mathrm{s}$, what fraction of neutrons will decay before they travel a distance of $10\mathrm{m}$ ?

(mass of neutron $=1.675\times {10}^{-27}\mathrm{Kg}$ )

Show Answer

Solution:

Let $\mathrm{m}$ and $\mathrm{v}$ denote the mass and speed of neutron. Let $\mathrm{N}$ be the initial number of neutrons in the beam.

Then ${\displaystyle \frac{1}{2}}{\mathrm{mv}}^2=0.0327\mathrm{eV}=0.0327\times 1.6\times {10}^{-19}\mathrm{J}$

$$\therefore \nu ={\left[{\displaystyle \frac{2\times 3.27\times 1.6\times {10}^{-21}}{1675\times {10}^{-27}}}\right]}^{1/2}{\mathrm{ms}}^{-1}=2.5\times {10}^3{\mathrm{ms}}^{-1}$$

The time dt taken by neutrons to travel a distance of $10\mathrm{m}$ is

$$\mathrm{dt}={\displaystyle \frac{10}{2.5\times {10}^3}}=4\times {10}^{-3}\mathrm{s}$$

Since dt is very small as compared to half life of neutrons

$$\begin{array}{rl}& {\displaystyle \frac{\mathrm{dN}}{\mathrm{N}}}=\lambda \mathrm{dt}\\ & \text{ or }{\displaystyle \frac{\mathrm{dN}}{\mathrm{N}}}={\displaystyle \frac{0.693}{700}}\times 4\times {10}^{-3}\\ & =8.96\times {10}^{-4}\end{array}$$

Mass Defect and Binding Energy

Very accurate measurements of mass of atomic nuclei showed that the experimental mass $(M)$ is slighly less than the sum of masses of nucleons $\left({\mathrm{M}}^{\prime }\right)$ in free state;

$${\mathrm{M}}^{\prime }={\mathrm{Zm}}_{\mathrm{p}}+(\mathrm{A}-\mathrm{Z}){\mathrm{m}}_{\mathrm{n}}$$

${\mathrm{m}}_{\mathrm{p}}$ and ${\mathrm{m}}_{\mathrm{n}}$ denote mass of proton and neutron respectively. The difference between the two masses in known as the mass defect $(\mathrm{\Delta }\mathrm{M})$, i.e.

$\mathrm{\Delta }\mathrm{M}=$ Mass defect $={\mathrm{M}}^{\prime }-\mathrm{M}$

An important question that arises is : what happens to mass $\mathrm{\Delta }\mathrm{M}$ ? According to our present knowledge let us consider that seperate neutrons and protons are availabe. We form a nucleus by putting appropriate number of neutrons and proton within the size of the nucleus. In the process work has to be done to overcome replusive force protons exert on one another. Work is nothing but energy. Where from this energy comes to form a nucleus? We consider that the required energy comes from the system itself. In terms of mass-energy equivalence this energy is due to conversion of mass $\mathrm{\Delta }\mathrm{M}$ into energy. This energy is known as the binding energy of the nucleus. Therefore

Binding energy $=$ mass defect $\times {\mathrm{c}}^2=(\mathrm{\Delta }\mathrm{M})\times {\mathrm{c}}^2$

We can also say an energy equivalent to the binding energy will have to be given to break a nucleus into its constituent nucleous lying very far off from one another.

Binding Energy of A Deutron

Mass of a proton $=1.007276\mathrm{amu}$

Mass of a neutron $=1.008665\mathrm{amu}$

Mass of a proton + Mass of a neutron $=2.015941\mathrm{amu}$

Experimental mass of a deutron nucleus $=2.013553\mathrm{amu}$

$\mathrm{\Delta }\mathrm{M}=($ the mass defect $)=0.002388\mathrm{amu}$

Binding Energy $=0.002388\times 931\approx 2.2\mathrm{MeV}$

The above result has been verified from a direct experiment. On bombarding a hydrogen nucleus with neutron, a neutron is captured in accordance with the reaction.

$${}_0{\mathrm{n}}^1+{}_1{\mathrm{H}}^1\to {}_1{\mathrm{H}}^2+\gamma$$

The energy of emitted $\gamma$ ray truns out to be $2.2\mathrm{MeV}$ in accordance with above predication. Also to carry out reverse reaction

$${}_1{\mathrm{H}}^2+\gamma \to {}_1{\mathrm{H}}^1+{}_0{\mathrm{n}}^1$$

The energy released has to be $2.2\mathrm{MeV}$

Stability of Nucleus and Binding Energy

A great deal of information about nuclei and their stability can be obtained from the studies of binding energy. However the calculation are in general, made for atoms rather than atomic nuclei. Calculation for binding energy per nucleon has been worked out for the various atoms in the periodic table. Figure shows the plot of binding energy per nuleon against the mass number. From the graph the following sailent features are observed.

(1) As a first approximation binding energy per nucleon starts with a low value of deutron and than increases rapidly.

(2) Some nuclei like ${}_2{\mathrm{He}}^4,{}_6{\mathrm{C}}^{12}$ and ${\mathrm{O}}^{16}$ have exceptionally high values as compared with their neighbours.

(3) After the rapid increase the binding energy per nucleon is found to decrease slowly as the number of nucleons increases beyond 60 .

(4) The nuclei in the region of the low to middle of peridic table are most tigetly bound. For example iron cobalt, nickel belong to this catagery.

(5) The analysis of curve shows that higher is the binding energy per nucleon, the more stable is the atom and vice versa.

The graph also gives us a very vital information. The shape of this average binding energy curve indicate that there are two general processes which can release energy fromnuclei.

(a) Combining light nuclei into a more massive nucleus.

(b) Spilting up heavy nuclei of into nuclei medium mass.

This is because in both cases, the total binding energy of the products exceeds that of reactants. These two cases corresponds to phenomenon of nuclear fission and nuclear fusion.

Example-14:

Calculate the energy liberated when a single helium nucleus is formed by the fusion of two deuterium nuclei. Given

Mass of ${\mathbf{H}}^2=\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{1}\mathbf{4}\mathbf{7}\mathbf{8}\mathrm{u}$

Mass of ${}_2{\mathrm{He}}^4=4.00388\mathrm{u}$

Show Answer

Solution:

The reaction can be written as

$${}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^2\to {}_2{\mathrm{He}}^4+\text{ Energy }$$

Mass defect $\mathrm{AM}=[2\times 2.01478-4.00388]\mathrm{amu}=0.02568\mathrm{u}$

Hence energy released is $=0.2568\times 931\mathrm{MeV}\approx 24\mathrm{MeV}$

Nuclear Fission

When a uranium nucleus was bombared by neutrons, it was found that the products of the reaction were having a number of different half lives. It was assumed that in these reactions, we are producing trans uranium elements. If we attribute these different half lives to the successive disintegration of the same atom, one can explain the possibility of a single nucleus emitting several $\beta$-particles one after another. Obviousely each $\beta$ emission increases the atomic number of resultant nucleus by one and a series of new transuranium elements can be stipulated. These result aroused tremendous interest in the physical world.

O. Hahn and F. Strassmann then tried to identify the products of the reaction when uranium was barbarded will neutrons chemically and found to surprise of every one that these products were a number of different elements nearly all of which are in the centre of periodic table. They thus concluded that the uranium nucleus after the capture of a neutron seemed to be spilting apart into two nearly equal fragments as illustrated in the figure. These results were supported by Miss Lise Meitner and S. Frisch. They suggested that the neutron initiated a decomposition of the uranium nucleus into two nuclei of roughly equal size. The phenomenon was termed as nuclear fission.

Fission

In general, fission fragments are not stable but contain an excess number of neutrons. Hence they either ejects neutrons or convert them into protons by ejecting $\beta$-particles. We list below some examples

2.

Further experiment showed that uranium after neutron capture divides itself into many possible fragments each corresponding to elements near the centre of periodic table. We list below some of the possible fragment pairs.

1. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to {}_{56}^{144}\mathrm{Ba}+{}_{36}^{89}\mathrm{Kr}+3{}_0^1\mathrm{n}+\mathrm{Q}$

2. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to \underset{\text{ Antimoney }}{{}_{51}^{133}\mathrm{Sb}}+{}_{41}^{92}\mathrm{Nb}+3{}_0^1\mathrm{n}+\mathrm{Q}$

3. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to {}_{\text{Xenon }}^{{}_{54}^{140}\mathrm{Xe}}+{}_{\text{Strontium }}^{{}_{83}^{94}\mathrm{Sr}}+2{}_0^1\mathrm{n}+\mathrm{Q}$

4. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to {}_{\text{Cadmium }}^{{}_{98}^{118}\mathrm{Cd}}+{}_{44}^{115}\mathrm{Ru}$

5. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to {}_{\text{Silver }}^{{}_{47}^{113}\mathrm{Ag}}+\underset{\text{ Rhodium }}{{}_{45}^{110}\mathrm{Ru}}+2{}_0^1\mathrm{n}+\mathrm{Q}$

6. ${}_{92}^{235}\mathrm{U}+{}_0^1\mathrm{n}\to \underset{\text{ Bromine }}{{}_{35}^{87}\mathrm{Br}}+\underset{\text{ Lanthanum }}{{}_{57}^{146}\mathrm{La}}+3{}_0^1\mathrm{n}+\mathrm{Q}$

$\mathrm{Q}$ in all the above reactions represents the energy released in the fission reaction. The value of $Q$ in the fission reaction is about $200\mathrm{MeV}$. This energy release from a single atom is formidable contribution. Hence the energy release can be enormous if a very large number of atoms undergo fission.

This is the underlying principal of a fission bomb. We illustrate a chain reaction in the following diagram.

The explosive release of nuclear energy by fission is used in the design of bombs. In a bomb, the material is chosen and arranged in such a way that each neutron has the greatest possible probability of producing another fission. The arrangment ensures that on the average one fission leads to more than one subsequent fission, a rapidly increasing number of fissions occur each acompained by the release of about $200\mathrm{MeV}$ energy. This heats the bomb and causes an explosion of infamous violence.

Nuclear Reactor

We design a nuclear reactor on the basis of a controlled fission reaction by the arrangement of reactor material. A steady reaction is maintained, in which each fission produces on an average of one new fission. We try to achieve a critical balance for the production of controlled nuclear energy.

The fission of uranium by neutrons is self sustaining chain reaction in which 2.5 neutrons per fission are produced on the average. The reaction can therefore proceed in a sustained and controlled way if one of these neutrons is able to cause further fission and remaining 1.5 neutron are allowed to escape. To meet these requireements the nuclear reactor is made sufficiently large to increase the (surface / volume) to the desired optimum level. Further if need be the fuel is also surrounded by a neutron reflecting material. This helps to decrease the ’neutron loss’ if it is sufficiently fast. The essential parts of a nuclear reactor are shown in the figure.

1. Fissionable fuel (like enriched uranium) taken usually in the form of rods.

2. Control rods are made of some suitable neutron absorbing material like cadium. They are pushed in and out to control the rate of neutron absorption and hence the rate of further fission.

3. Moderator (like graphile, heavy water) that helps to slow down the fast neutrons and thus increasing the probability of their capture by uranium. It shares the energy of the fast moving nuetrons without actually absorbing them. The moderator is prefeably a material whose atomic mass is comparable to that of a neutron and which therefore causes a large energy exchange on each collision. The moderator is suitably mixed with the uranium fuel.

The nuclear reactor goes into operation when some stray neutrons initiate fission in the enriched uranium fuel. The moderator and control rods are then suitably adjused to sustain the chain reaction at a controlled rate. The energy released is used to produce steam in a steam generator. The steam then drives the turbines connected to the electric generator and thus help to produce electricity.

Nuclear Fusion

Nuclear energy can also possibly be obtained by the fusion process of light nuclei. The basic underlying principle to obtain fusion nuclear energy is to fuse two or more light nuclei of combined mass ’ $m$ ’ to produce a single nucleus of mass less than $\mathrm{m}$.

Some typical fusion reactions had been produced in the labortary by bombarding (of appropriate targets) particles, accelerated to very high energies with suitable particle accerlators. Some typical fusion reactions are listed below:

$$\begin{array}{rl}& {}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^2\to {}_1{\mathrm{H}}^3+4\mathrm{MeV}\\ & {}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^2\to {}_2{\mathrm{He}}^3+{}_0{\mathrm{n}}^1+3.3\mathrm{MeV}\\ & {}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^3\to {}_2{\mathrm{He}}^4+{}_0{\mathrm{n}}^1+17.5\mathrm{MeV}\\ & {}_1{\mathrm{H}}^2+{}_2{\mathrm{He}}^3\to {}_2{\mathrm{He}}^4+{}_1{\mathrm{H}}^1+{}_1{\mathrm{H}}^1+18.3\mathrm{MeV}\end{array}$$

One main reason the fusion energy release is becoming significantly important is the fact that theoretically there is a large scale possibility of obtaining energies from the fusion of deuterium and deuterium can be obtain in abudence at a significantlly low cost from the sea water. However there are numerous practical difficulties which are to be overcome before using fusion reactions as a source of nuclear energy.

The first and forcemost difficulty in bringing about fusion is that the nuclei to be fused are usually positive charged and so they tend to push apart because of very strong electrostatic repulsive forces. The only way out to overcome this difficulty is that nuclei must be raised to excessively high temperatures (estimated to be about 100 million degree). At such high temperature the random internal motion of particles would bring them within the range of nuclear forces for fusion to occur. This has been achieved in an uncontrolled way in so called fusion bomb.

However a fusion reactor is a distant future possiblity. First we have to develop techniques to achieve fantastic temperatures of 100 million degrees. Then we have to keep fusing nuclei in that region long enough to produce enough energy from the reaction to make it self sustained. The most serious difficulty is however to find suitable containers to confine the material while it is reacting at such high temperatures. Developments in Plasma Physcis has developed so called Magnetic Bottles. At temperature of 100 million degrees the nuclei of fusing nuclei are in an ionized state and behave like a swirring mass of electric charges. One can therefore hope that the under influence of specially designed configuration magnetic fields or electric fields, the reacting material in the form of “Plasma” can be confined within the field bounderies. Magnetic bottles can force the plasma to stay in the region bounded only by magnetic lines of force.

Fusion reactors, (when fully developed) will have the following distinct advantages viz. the nuclear fuel is usually obtained from water which would always be available. More over the products are free from radioactive waste and quite harmless.

Nuclear Forces

The three basic characterstics of nuclear forces are as follows:

1. Nuclear forces are very strong attractive forces.

2. Nuclear forces are short range forces.

3. Heavy nuclei for their stability must also have a repulsive component such that the nuclear forces are on the whole attractive.

We now proceed to discuss these points one by one.

Our studies of nuclear reactions shows that the energies involved in unclear reactions are tremendously large as compared to the corresponding energy changes involved in chemical reaction. This shows that the force which are responsible for holding of its nucleons must be very strong. Another clue about the fact that the nuclear forces are very strong is obtained from the enormously large value of nuclear densities. The fact that atomic nuclei are stable and contain protons within nuclear volumes provide another support to the strong nature of nuclear forces. The presence of protons within nucleus must involve enormously large electrostatic repulsive force. The only way one can explain the stability of the nucleus, is to assume the existance of very strong attractive forces between nucleons. These strong (and short range) nuclear forces must overcome the electrostatic repulsive forces. We have seen that the binding energy per nucleon (exception light nuclei) has the same value $(\sim 8\mathrm{MeV})$ for all nuclei. Thus we can conclude that binding energy of a nucleus is proportional the number of nucleons constituting the nucleus. Assuming that every nucleon in a nucleus of mass number A interacts with every other nucleon, the number of interacting pairs

would be ${\displaystyle \frac{A(A-1)}{2}}$. Hence the total interaction energy and therefore the binding energy would the proportinoal to ${\mathrm{A}}^2$. However this is not the case in the least.

A clear deduction is that every nucleon in the nucleus does not interact with every other nucleon. Hence a nucleon is only intteracting with its immediate nearest neighbour. Thus nuclear forces must have a short range. Nuclear forces falls of very rapidly with the distance between two nucleons, the decrease has to be much more rapid than ${\displaystyle \frac{1}{{\mathrm{r}}^2}}$.

Experimental studies show that if the nuclear forces holding the nucleons were entirely attractive, heavy nuclei would collapse. This is so because in heavy nuclei, the nucleon’s are so closed logether so that each one is within the range of attractive forces of every other nulceon. This would ultimately mean that within the same nuclear dimensions the attractive energy acting on each particle would grow with the total number of nucleons. This contradicts the facts observed in reality. Thus we all are led to unescaple conclusion that nuclear force must also have a repulsive component that saves the nuclei from collapsing. Thus nuclear forces only on the balance are attractive.

Example-15:

Calculate the force between two fission fragments of equal masses and sizes that are produced by the fission of ${\mathrm{Pu}}^{239}$ by a thermal neutron in which four neutrons are emitted.

Show Answer

Solution:

Total mass number of ${}_{94}{\mathrm{Pu}}^{239}$ and a neutron $=246$

Mass number of each fragment $={\displaystyle \frac{240-4}{2}}=118$

Charge on each fragment $={\displaystyle \frac{94}{2}}=47$

Radius of nucleus so formed $=1.2\times {10}^{-14}{\mathrm{A}}^{1/3}$ where $\mathrm{A}$ is the mass number

$$\therefore \mathrm{r}=1.2\times {10}^{-15}\times (118{)}^{1/3}=5.886\times {10}^{-15}$$

Distance between the centres of two fragments $=2\times 5.886\times {10}^{-14}=11.77\times {10}^{-15}$

$\therefore$ Force between two fragment ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}{\displaystyle \frac{47\times 47}{{(11.77\times {10}^{-15})}^2}}$

$=3.763\times {10}^2\mathrm{N}$

Example-16:

A nuclear explosion is required to deliver $1\mathrm{MeV}$ of heat energy. How many fission events must be required to attain this power level. If this explosion is designed with a nuclear fuel consisting of ${\mathrm{U}}^{235}$ to run a reactor at this power level for one year then calculate the amount of fuel needed. It is given that energy released per fission event is $200\mathrm{MeV}$.

Show Answer

Solution:

Energy required by nuclear explosion $=1\mathrm{MW}={10}^6{\mathrm{Js}}^{-1}$

Energy released by fission $=200\mathrm{MeV}$

$$=200\times 1.6\times {10}^{-13}=3.2\times {10}^{-11}\mathrm{J}$$

No. of fission required per second $={\displaystyle \frac{{10}^6}{3.2\times {10}^{-11}}}=3.125\times {10}^6$

Number of fission required per year by $1\mathrm{MW}$ energy release

$$=3.125\times {10}^{16}\times 365\times 24\times 60\times 60=9.9\times {10}^{23}$$

Hence amount of uranium fuel needed is $={\displaystyle \frac{9.9\times {10}^{23}\times 235}{6.02\times {10}^{23}}}$

$$=386.46\mathrm{g}$$

Example-17:

$20\mathrm{MeV}$ energy is released per fusion of reaction

$${}_1{\mathbf{H}}^2+{}_1{\mathrm{H}}^2\to {}_2{\mathbf{H}\mathbf{e}}^4+{}_0{\mathbf{n}}^1$$

Calculate the mass of ${\mathrm{H}}^2$ consumed in a fusion reaction of power $10\mathrm{MW}$ in 100 days.

Show Answer

Solution:

$10\mathrm{MW}$ of power is to be produced by the fusion reactor for 100 days which means that $10\times {10}^6={10}^7$ energy is required in $100\times 24\times 60\times 60=8640000\mathrm{s}$.

Now energy released per fusion is $20\mathrm{MeV}=20\times 1.6\times {10}^{-13}=3.2\times {10}^{-12}\mathrm{J}$

Mass of ${}_1{\mathrm{H}}^2$ consumed in one fusion $=4\mathrm{amu}=4\times 1.66\times {10}^{-24}\mathrm{Kg}=6.64\times {10}^{-27}\mathrm{K}$

Hence required mass ${}_1{\mathrm{H}}^2$ is given by ${\displaystyle \frac{6.64\times {10}^{-27}}{3.2\times {10}^{-12}}}\times 8640000\times {10}^6$

$$=17.9\times {10}^{-4}\mathrm{kg}$$

Example-18:

The volume of water in the Great Lakes is $\mathbf{5}\mathbf{0}\mathbf{0}\mathbf{0}$ cubic mile. Calculate the amount of energy released if all the deutorium atoms in water are used up in fusion. Given that the abudence of deutorium is $0.0156\mathrm{\%}$ and the fusion of 6 deutorium atoms gives an energy of $43\mathrm{MeV}$. Given $1\mathrm{mile}=1.6\mathrm{Km}$

Show Answer

Solution:

The volume of water in the Great Lakes $=5000$ cubic mile

The mass of water $=5000\times {(1.6\times {10}^3)}^3\times {10}^3\mathrm{Kg}=2.13\times {10}^{16}\mathrm{kg}$

Since molecular weight of water is 18 , the total number of molecules in water having a mass of $2.13\times {10}^{16}\mathrm{kg}$

$$={\displaystyle \frac{2.13\times {10}^{16}\times 6.023\times {10}^{23}}{18}}=7.13\times {10}^{41}$$

Since the abudance of deutorium is $0.0156\mathrm{\%}$, we have total number of deutorium atoms is $=7.13\times {10}^{41}\times 2\times 0.0156\times {10}^{16}=2.22\times {10}^{28}$

Now the fusion of 6 deutorium atoms gives an energy of $43\mathrm{MeV}$, therefore the energy release per atom is

$${\displaystyle \frac{43}{6}}=7.17\mathrm{MeV}$$

Hence the total release of energy of $2.22\times {10}^{28}\times 7.17\mathrm{MeV}=1.6\times {10}^{29}\mathrm{MeV}$

PROBLEMS FOR PRACTICE

1. With what energy an $\alpha$ particle be directed towards the gold nucleus $(\mathrm{z}=79)$ so that if may retrace its path from a point which is $3\times {10}^{-14}\mathrm{m}$ away from the gold nucleus?

Show Answer

[Answer: $6.9\mathrm{MeV}$ ]

2. The number of alpha particle scattered at ${60}^{\circ }$ is 100 per minute in an $\alpha$ particle scattering experiment. What would be the number of alpha particles scattered per minute at ${90}^{\circ }$.

Show Answer

[Answer: 25]

3. If the first member of the Lyman’s series is $1215\text{AA}$ calculate the wavelength of the first member of Paschen and Brackett series

Show Answer

[Answer: $18752\text{AA},54000\times {10}^{-18}\mathrm{m}$ ]

4. The wave number of a spectral series are 4857, 19932, 25215, 27663 and 27997 show that the limit of this seires is approximately 32033 .

5. A $10\mathrm{kg}$ satellite circles the earth every 2 hour in an orbit having a radius of $8000\mathrm{kg}$. Assuming that the Bohr’s angular momentum applies to the satellite as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite.

Show Answer

[Answer: $5.18\times {10}^{45}$ ]

6. Which level of the doubly ionized litium has the same energy as the ground state energy of the hydrogen atom.

Show Answer

[Answer: $3^{\text{rd }}$ level]

7. The energy needed to detach the electron of a hydrogen like ion in ground state is equal to $4\mathrm{R}$. What is the wavelength of the radiation emitted with a transition from first excited state to a ground state?

Show Answer

[Answer: $30.4\mathrm{nm}$ ]

8. A positive ion having just one electron ejects it when a photon of wavelength $228\times {10}^{-10}\mathrm{m}$ or less is absorbed by it. Name the ion

Show Answer

[Answer: ${\mathrm{He}}^{+}$]

9. What is the energy of a hydrogen atom in the first exicted state if the potential energy is taken to zero in the ground state.

Show Answer

[Answer: $23.8\mathrm{eV}$ ]

10. Calculate the weight of $\mathrm{RaB}$ showing radioactivity of 1 curie.

Show Answer

[Answer: $3.1\times {10}^8\mathrm{g}$ ]

11. Calculate the time required for $10$ of a sample of thorium to disintegrate. Assume the half-life of thorium to be $1.4\times {10}^{10}$ year.

Show Answer

[Answer: $2.1\times {10}^9$ year]

12. 10 milligram of a radioactive material of half life period of two years are kept in a store for four years. How much of the matter remain unchanged.

Show Answer

[Answer: $2.501\mathrm{mg}$ ]

13. One milligram of thorium emits $22\alpha$ particle per unit solid angle per minute. Calculate the half life of thorium.

Show Answer

[Answer: $1236\times {10}^{10}$ year]

14. The half life of radium is 1500 year. In how many year will $1\mathrm{g}$ of radium (a) lose $1\mathrm{mg}$ (b) be reduced to one centigram.

Show Answer

[Answer: 1.995 year; $9.972\times {10}^3$ year]

15. The activity, of a radioactive sample falls from $400{\mathrm{s}}^{-1}$ to $200{\mathrm{s}}^{-1}$ in a minute. Calculate the half life of radioactive sample.

Show Answer

[Answer: 40 minute]

16. For every ${10}^6$ atoms of radium in a sample today, find the number of radium atoms that will be left after 3200 years. Half life of radium $=1600$ year.

Show Answer

[Answer: $2.5\times {10}^5$ atom]

17. A certain radioaction substance has a half life of 5 years. What will be the probability of decay of this substance in 10 years.

Show Answer

[Answer: 75 %]

18. Plutonium decays by the following reaction with a half life of 2400 years

$${}_{94}{\mathrm{P}}^{239}\to {}_{92}{\mathrm{U}}^{235}+{}_2{\mathrm{He}}^4$$

If Plutonium is stored for 7200 years what fraction of it remains?

Show Answer

[Answer: $1/8$ ]

19. An isolated rock was found to contain ${\mathrm{m}}_1\mathrm{g}$ of ${\mathrm{U}}^{238}$ and ${\mathrm{m}}_2\mathrm{g}$ of ${\mathrm{Pb}}^{206}$. If the half life of ${\mathrm{U}}^{235}$ is $4.6\times {10}^9$ years calculate the age of the rock.

Show Answer

[Answer: $\left[{\displaystyle \frac{\mathrm{T}\log (1+{\displaystyle \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}}\times {\displaystyle \frac{238}{206}})}{\log 2}}\right]$]

20. Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{78}{\mathrm{Au}}^{197}$ and silver isotope ${}_{47}{\mathrm{Ag}}^{107}$. What is the approximate ratio of their nuclear mass densities.

Show Answer

[Answer: 1.23, of the order of one]

21. The total binding energy of ${}_{10}^{20}\mathrm{N}$ is $160.6\mathrm{MeV}$. Find its atomic mass.

Show Answer

[Answer: $19.19\mathrm{u}$ ]

22. Find the mass of ${}_6^{14}\mathrm{C}$ in the following reaction

$${}_7^{14}\mathrm{N}+{}_0{\mathrm{n}}^1\to {}_6^{14}\mathrm{C}+{}_1^1\mathrm{H}+0.55\mathrm{MeV}$$

The atomic mass of ${}_7^{14}\mathrm{N},{}_0^1{\mathrm{n}}^{+}$and ${}_1^1\mathrm{H}$ are $14.03074,1.008665$ and 1.007825 respecitvely.

Show Answer

[Answer: 14.003323]

23. Calculate the energy required to remove an al pha particle from $0^{16}$ to produce $C^{12}$. Binding energy per nuleon.

$${}^{16}\mathrm{O}=7.97\mathrm{MeV},{}^{12}\mathrm{C}=7.68\mathrm{MeV},{}^4\mathrm{He}=7.97\mathrm{MeV}$$

Show Answer

[Answer: $7.12\mathrm{MeV}$ ]

24. Calculate the energy released when a thermal neutron enters the nucleus of ${\mathrm{Pu}}^{239}$. Given masses of ${\mathrm{Pu}}^{239}$ and ${\mathrm{Pu}}^{240}$ are 239.1270 amu and 240.1291 amu.

Show Answer

[Answer: $6.349\mathrm{MeV}$ ]

25. Find the binding energy per nucleon of ${}_3^7\mathrm{Li}$. Given mass of proton $=1.00782\mathrm{amu}$, mass of a neutron $=1.00866\mathrm{amu}$ and mass of ${}_3^7\mathrm{Li}=7.01599\mathrm{amu}$.

Show Answer

[Answer: $5.6\mathrm{MeV}$ ]

26. The masses of ${}^{11}\mathrm{C}$ and ${}^{11}\mathrm{B}$ are respectively $11.011\mathrm{u}$ and $11.0093\mathrm{u}$. Find the maximum energy of a poistron can have in the $\beta$ decay of ${\mathrm{C}}^{11}$ to ${\mathrm{B}}^{11}$.

Show Answer

[Answer: $933.6\mathrm{KeV}$ ]

27. How much energy will be produced on complete fission of one gram of uranum taking that only $0.1\mathrm{\%}$ of mass is converted into energy.

Show Answer

[Answer: $8.97\times {10}^7\mathrm{J}$]

28. Calculate the power output of ${}_{92}^{235}\mathrm{U}$ reactor if it takes one month to use up $2\mathrm{kg}$ of uranium fuel.

Show Answer

[Answer: $62.5\mathrm{MeV}$ ]

29. The fission of ${}_{92}^{235}\mathrm{U}$ produces a mean energy of $185\mathrm{MeV}$. If ${}_{92}^{235}\mathrm{U}$ in a reactor is continuously genrating $100\mathrm{MeV}$ of power how long will take for $1\mathrm{kg}$ of uranium to be used up?

Show Answer

[Answer: 8.78 day]

30. In the fission of a single nucleus of ${}_{92}^{235}\mathrm{U},200\mathrm{MeV}$ energy is released. What is the number of fissions which must occur per second so as to generate a power of $4\mathrm{kW}$.

Show Answer

[Answer: $12.5\times {10}^{13}$ ]

31. When a atom of ${\mathrm{U}}^{235}$ undergoes fission in a reaction about $200\mathrm{MeV}$ energy is released. Suppose that a reactor using ${\mathrm{U}}^{235}$ has an ouput of $70\mathrm{MW}$ and is $20\mathrm{\%}$ efficient. How many uranium atoms does it consume in one day? What mass of uranium does it consume each day?

Show Answer

[Answer: $9.5\times {10}^{24}$ and $3.7\mathrm{kg}$ ]

32. Calculate the energy released in the fusion reaction

$${}_2^4\mathrm{He}+{}_2^4\mathrm{He}+{}_2^4\mathrm{He}\to {}_6^{12}\mathrm{C}$$

Given the mass of an $\alpha$ particle is $4.002603\mathrm{amu}$

Show Answer

[Answer: 25]

33. Calculate the amount of energy released when two deuterium nuclei fuse into an $\alpha$ particle.

Mass of deuterium $=2.014102\mathrm{amu}$ and mass of $\alpha$ particle $=4.00263\mathrm{amu}$

Show Answer

[Answer: $23.8\mathrm{MeV}$ ]

34. It is proposed to use nuclear fusion reaction

$${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\to {}_2^4\mathrm{He}$$

in a nuclear reactor of $200\mathrm{MW}$ rating. If the energy from above reaction is used with a $25\mathrm{\%}$ efficiency in the reactor, how many grams of deuterium will be needed per day? Given masses of, ${}_1{\mathrm{H}}^2$ and ${}_2{\mathrm{H}}^4$ are $2.0141\mathrm{u}$ and $4.0026\mathrm{u}$ respectively.

Show Answer

[Answer: 121.39]

35. Calculate the energy that can be obtained from $1\mathrm{kg}$ of water through fusion reaction.

$${}^2\mathrm{H}+{}^2\mathrm{H}\to 3\mathrm{H}+\mathrm{p}$$

It is given that of natural water is heavy water (by number of molecules) and all the deuterium is used for fusion.

Show Answer

[Answer:: 3200 MJ]

36. The count rate from a radioactive sample falls from $4.0\times {10}^6$ per second to $1.0\times {10}^6$ per second in 20 hours. What will be the count rate 100 hours after the begining?

Show Answer

[Answer: $2.8\times {10}^6$ distegrations per second]

37. The decay constant of ${\mathrm{U}}^{238}$ is $4.9\times {10}^{-18}{\mathrm{s}}^{-1}$. By what factor does the activity of a ${\mathrm{U}}^{238}$ decrease in $9\times {10}^9$ years.

Show Answer

[Answer: 4]

Question Bank

Key Learning Points

1. The discovery of the electron and the fact that electron is a common constituent of all matter established the fact that atom has a structure.

2. Any model of the atomic structure must explain the stability and the spectra of atoms.

3. According to Thomson’s model, the atoms consist of positively charged entities distributed uniformly over the entire body of the atom with negative electrons embeded in this continuous positive charge.

4. Geriger and Mardsen in their classic experiments on scattering of $\alpha$ particles found that most of the $\alpha$ particles emerged without deviation but some of them were scattered through very large angles.

5. Rutherford concluded from the results of Gerger and Madsen experiment that atoms contain a tiny nucleus in which positive charge and nearly all of its mass is concentrated.

6. The electrons in Rutherford model are assumed to be moving around the nucleus in circular orbits.

7. Rutherford atomic model could not explain the stability of the atoms when subjected to the demand of classical electromagnetic theory that any accelerated charged particle must emit energy in the form of electromagnetic radiation.

8. Bohr modified Rutherrod atomic model by introducing quantisation of angular momentum of electron orbits in which electron can revolve around the nucleus.

9. According to first postulate of Bohr’s theory the electron can move only in certain privileged orbits which are not sujbected to the demand of electromagnetic theory. These privileged orbits known as stationary orbits are the ones in which the angular momentum of the electron is an integral multiple of ${\displaystyle \frac{\mathrm{h}}{2\pi }}$.

10. According to Bohr’s theory any emission or absorption of radiation correspond to a transition between two stationary orbits.

If an electron transition take place from an allowed orbit with energy ${\mathrm{E}}_1$ to an allowed orbit with energy ${\mathrm{E}}_2$ the frequency of the emitted radiation is given by the relation

$$\mathrm{h}\nu ={\mathrm{E}}_1-{\mathrm{E}}_2$$

The energy of an electron with quantum number $n$ is given by ${\mathrm{E}}_{\mathrm{n}}=-{\displaystyle \frac{13.6}{{\mathrm{n}}^2}}\mathrm{eV}$

11. Bohr’s theory could successfully explain the spectra of various series in the hydrogen and hydrogen like atoms. However Bohr’s theory could not explain the fine structure of spectral lines.

12. The phenomenon of radioactivity is defined as the spontaneous disintegration of nuclei (with atomic number greater than that of lead) of one or more atoms with the emission of highy penetrating radiations.

13. The radioactive radiation consist of three distinct parts known as $\alpha$ rays, $\beta$ rays and $\gamma$ rays. $\alpha$ particles are nothing but doubly ionized helium nuclei.

$\beta$ particles are negatively charged electrons moving with speeds comparable with the velocity of light.

$\gamma$ rays are electromagnetic waves of very short wavelength.

14. According to Rutherford and Soddy theory whenever a radio disintegration occurs, it does so with the emission of a $\alpha$ particle or a $\beta$ particle.

15. In radioactive transformations, it is observed that the daughter from a given parent is also usually radioactive. The daughter nucleus changes to a new nucleus by an $\alpha$ or $\beta$ emission. The chain continue until a stable element usually a stable isotope of lead is reached. Such a change of radioactive transformation consitutues a radioactive series.

16. Radioactive series occuring in nature are the uranium series, the actinum series, the thorium series and the neptunium series.

17. Radioactive disintegrations are governed by the laws of probabilily.

18. The mathematical relation representing the radioactive disintegration is expressed by the relation

$$\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\lambda \mathrm{t}}$$

where ${\mathrm{N}}_0$ represents the number of radioactive atoms at $\mathrm{t}=0$ and $\mathrm{N}$ represents the number of radioactive atoms at a later instant $t$. $\lambda$ is known as the disintegration constant.

19. Radioactivity in a radioactive element completely disappear only in an infinite time and all radioactive elements are identical in this respect.

20. To distinguish the activity of one radioactive element from another we introduce the concept of half and mean life.

21. The half life of a radioactive element is defined as the time in which the number of radioactive atoms is reduced to half of its initial value. The half life $\mathrm{T}$ is given by the relation: $\mathrm{T}={\displaystyle \frac{0.693}{\lambda }}$.

22. The mean life is defined as the ratio of combined age of all the atoms and the number of atoms in the concerned sample. Mean life is given by the relation. $\tau ={\displaystyle \frac{1}{\lambda }}$.

23. A curie is equal to $3.7\times {10}^{10}$ disintegration per second.

24. Radio isotopes can be used in tracer technique in agriculture, in industry and in radio carbon dating.

25. According to Rutherford atom has a small positively charged nucleus. Practically the whole mass of the atoms is concentrated in the nucleus.

The nucleus consists of protons and neutrons. Neutron were discovered by Chadwick.

26. It is found that radius of a nucleus varies as cube root of its mass number. It is found that very approximately

$$r=r_0{A}^{1/3}$$

where $r_0$ is the distance of nearest approach to the nucleus. $r_0$ is referred to as nuclear unit radius and is nearly equal to $1.4\mathrm{fm}$.

27. The energy changes encountered in nuclear reactions are due to conversion of mass into energy in accordance with Einstein relation

$$\mathrm{E}={\mathrm{mc}}^2$$

28. A detailed survey of all atoms reveal that atomic mass is always less than the sum of the masses of the constituent particles in free state. Thus the formation of an atom from its constituent particles always results in the appearance of mass defect.

29. The energy needed to break a nucleus into its constituent particles is known as binding energy.

30. The anlysis of the studies of binding energy shows that higher is the binding energy per nucleon the more stable is the atom and vice versa.

31. When uranium was bombarded with neutrons Fermi found that a compound nucleus is formed due to capture of neutron along with the emission of $\beta$ particles with a number of different half lives. He interpreted the phenomenon as the formation of trans uranium elements.

32. Hahn and Strassman after a series of careful chemical analysis of the products formed when uranium was bombarded with neutrons established that uramum nucleus after the capture of neutron seemed to be spliting apart into two nearly equal fragments.

33. Lise Meitner and R. Frisch suggested that neutrons iniated a decomposition of the uranium nucleus into two nuclei of roughly equal size. The phenomenon was termed as nuclear fission.

34. Due to fission a uranium nucleus split up into a large number of pairs of fission fragments with an average number of 2.5 nuetrons as the products of the reaction. The reaction is accomplished with a release of large amount of energy.

35. The average energy released per fission of uranium nucleus is of the order of $200\mathrm{MeV}$.

36. The neutron released in the fission process promotes further fission and set up a self sustained chain reaction.

37. The uncontrolled self sustained reaction can result in explosive release of energy. This is the underlying principle of fission bomb.

38. Nuclear reactor is designed to achieve a controlled release of energy. By the arrangment of the reactor material a study chain reaction is maintained in which each fission produces an average of one new fission.

39. Nuclear fusion is the phenonmenon in which two light nuclei are fused together to form a single new nucleus. These reactions are accompanied by the release of nuclear energy.

40. It is now believed that fusion process is primarly responsible for the release of energy by the sun and stars.

41. The forces that are responsible for holding of the nucleon together in a nucleus are very strong nuclear forces.

42. Nuclear forces are very stong short range attractive forces.

43. Studies of nuclei with higher values of atomic number reveal that in addition to very strong attractive interactions nuclear forces also have a repulsive component that saves the nuclei from collapsing.

Average

Bohr’s Model of the Atom

1. An electron makes a transition from an excited state to the ground state of a hydrogen like atom / ion. During this transition of the electron, its

(1) kinetic energy; potential energy and total energy all decrease

(2) kinetic energy decreases, potential energy increases and total energy remains same (Incorrect)

(3) kinetic energy and total energy decrease but potential energy increases

(4) kinetic energy increases, but potential energy and total energy both decrease

Show Answer

Correct answer: (4)

Solution:

In the given transition, principal quantum number (n) decreases. Now the speed of the electron in the ${\mathrm{n}}^{\text{th }}$ orbit, $\left(v_n\right)$, is inversly proportional to $n$. Hence, when $n$ decreases, $v_n$ and, therefore, the kinetic energy of the electron, increases. We also know

Total energy $=-\mathrm{K}.\mathrm{E}$

and Potential energy $=-2$ K.E

Due to -ve sign, in the expression for T.E and P.E; both these energies decrease when $\mathrm{n}$ decreases.

Hence option (4) is correct.

Average

Bohr’s Model of the Atom

2. Hydrogen $\left({}_1^1\mathrm{H}\right)$; deuterium $\left({}_1^2\mathrm{H}\right)$, singly ionised helium $\left({}_2^4\mathrm{H}\right)$ and doubly ionised lithium ${\left({}_2^8\mathrm{Li}\right)}^{++}$, all have only one electron orbiting around the nucleus. Consider an electron transition from $n=2$ to $n=1$. If the wavelengths emitted are ${\lambda }_1,{\lambda }_2,{\lambda }_3$ and ${\lambda }_4$, respectively, for these four atoms / ions. Then (approximately) which one of the followings is correct?

(1) $4{\lambda }_1=2{\lambda }_2=2{\lambda }_3={\lambda }_4$

(2) ${\lambda }_1=2{\lambda }_2=2{\lambda }_3={\lambda }_4$

(3) ${\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4$

(4) ${\lambda }_1=2{\lambda }_2=3{\lambda }_3=4{\lambda }_4$

Show Answer

Correct answer: (3)

Solution:

For a hydrogen like atom, of atomic number $z$; for the given transition, we have

$${\displaystyle \frac{1}{\lambda }}={\mathrm{Rz}}^2({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}})$$

Here $\mathrm{R}$ is Rydberg’s constant, for hydrogen. For the given four atoms ${\mathrm{z}}_1=1,{\mathrm{z}}_2=1,{\mathrm{z}}_3=2$, and ${\mathrm{z}}_4=3$. Therefore,

$$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_1}}:{\displaystyle \frac{1}{{\lambda }_2}}:{\displaystyle \frac{1}{{\lambda }_3}}:{\displaystyle \frac{1}{{\lambda }_4}}::1:1:(2{)}^2:(3{)}^2\\ & \therefore {\lambda }_1:{\lambda }_2:{\lambda }_3:{\lambda }_4::1:1:{\displaystyle \frac{1}{4}}:{\displaystyle \frac{1}{9}}\\ & \text{ or }{\lambda }_1={\lambda }_2=4{\lambda }_3=9{\lambda }_4\end{array}$$

Hence option(3) is correct.

Difficult

Bohr’s Model of the Atom

3. A photon colloides with a stationary hydrogen atom, in its ground state, inelastically. Energy of the colloiding photon is $10.2\mathrm{eV}$. After a time interval, of the order of a microsecond, another photon colloides with the same hydrogen atom, inelastically, with an energy of 15 eV. A detector observes:

(1) 2 photons of energy $10.2\mathrm{eV}$

(2) 2 photons of energy $10.4\mathrm{eV}$

(3) 1 photons of energy $10.2\mathrm{eV}$ and an electron of energy $1.4\mathrm{eV}$

(4) 1 photons of energy $10.2\mathrm{eV}$ and another photon of energy $1.4\mathrm{eV}$

Show Answer

Correct answer: (3)

Solution:

The hydrogen atom, in its ground state, absorbs the incident photon of energy $10.2\mathrm{eV}$ and goes to its first excited state.

$$(\because E_2-E_1=[-3.4-(-13.6)]\mathrm{eV}=10.2\mathrm{eV})$$

The atom remains in the excited state for a time of the order of ${10}^{-8}\mathrm{s}$. In one microsecond, it returns to its ground state, emitting a photon of energy $10.2\mathrm{eV}$. Now the second photon, of (incident) energy $15\mathrm{eV}$ undergoes an inelastic collision with the same hydrogen atom. Out of $15\mathrm{eV}$ of energy, the atom recieves $13.6\mathrm{eV}$ are absorbed to knock out the electron from its ground state. The electron becomes free, i.e., hydrogen atom is ionised. The remaining energy, $(15-13.6)=1.4\mathrm{eV}$ is the kinetic energy of the electron knocked out of the hydrogen atom.

Hence option (3) is correct.

Difficult

Bohr’s Model of the Atom

4. In a hypothetical hydrogen like atom; the electric potential between a proton and an electron, distant $r$ apart, is given by $V=V_0\ell n\left({\displaystyle \frac{r}{r_0}}\right)$

where $V_0$ and $r_0$ are constants. Assuming that the atom still obeys Bohr’s postulates, the dependence of the radius of the $n^{\text{th }}$ allowed orbit, $r_n$, on the principal quantum number $(n)$ is as

(1) $r_n\propto n$

(2) ${\mathrm{r}}_{\mathrm{n}}\propto {\mathrm{n}}^{-1}$

(3) $r_n\propto n^2$

(4) $r_n\propto n^{-2}$

Show Answer

Correct answer: (1)

Solution:

$\mathrm{U}=$ The electrostatic P.E of the electrons, at a distance r from the proton, $=\mathrm{eV}=\mathrm{e}{\mathrm{V}}_0\ell \mathrm{n}\left({\displaystyle \frac{\mathrm{r}}{{\mathrm{r}}_0}}\right)$

$\therefore \mathrm{F}=$ The magnitude of the force between the electron and proton $=|-{\displaystyle \frac{\mathrm{dU}}{\mathrm{dr}}}|={\displaystyle \frac{{\mathrm{eV}}_0{\mathrm{r}}_0}{\mathrm{r}}}$

The force, F, provides the centeripetal force for the electron, orbiting in its ${\mathrm{n}}^{\text{th }}$ allowed orbit of radius ${\mathrm{r}}_{\mathrm{n}}$.

$\therefore {\displaystyle \frac{{\mathrm{mv}}_{\mathrm{n}}^2}{{\mathrm{r}}_{\mathrm{n}}}}={\displaystyle \frac{{\mathrm{eV}}_0{\mathrm{r}}_0}{{\mathrm{r}}_{\mathrm{n}}}}$ $$ ……..(1)

From Bohr’s postulate, we also have ${\mathrm{mv}}_{\mathrm{n}}{\mathrm{r}}_{\mathrm{n}}=\mathrm{n}{\displaystyle \frac{\mathrm{h}}{2\pi }}$ $$ ……..(2)

From Eqn. (1) and (2), we get, ${\mathrm{r}}_{\mathrm{n}}\propto \mathrm{n}$

Hence option (1) is correct.

Difficult

Bohr’s Model of the Atom

5. A hydrogen atom, and a doubly ionised lithium $(\mathrm{Li}{)}^{++}$atom, are in their first excerted state and the second excited state, respectively. If ${\ell }_{\mathrm{H}}$ and ${\ell }_{\mathrm{Li}}$ are the respective angular momenta, and ${\mathrm{K}}_{\mathrm{H}},{\mathrm{K}}_{\mathrm{Li}}$ are the, respective, kinetic energies, of the electron, in the two atoms, we can say that

(1) ${\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{1}{2}};{\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\displaystyle \frac{4}{1}}$

(2) ${\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{2}{3}};{\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\displaystyle \frac{2}{3}}$

(3) ${\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{2}{3}};{\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\displaystyle \frac{1}{4}}$

(4) ${\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{1}{2}};{\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\displaystyle \frac{9}{4}}$

Show Answer

Correct answer: (3)

Solution:

For the hydrogen atom ${\mathrm{n}}_{\mathrm{H}}=2$ and ${\mathrm{z}}_{\mathrm{H}}=1$

For the doubly ionised lithium atom, we have ${\mathrm{n}}_{\mathrm{Li}}=3$; and ${\mathrm{z}}_{\mathrm{Li}}=3$

According to Bohr’s first postulate, the angular momentum of electron, in an orbit, is directly proportional to $\mathrm{n}$; therefore;

$$\equiv {\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{{\mathrm{n}}_{\mathrm{H}}}{{\mathrm{n}}_{\mathrm{Li}}}}={\displaystyle \frac{2}{3}}$$

The K.E of the electron, is an orbit, equals the |T.E. $\mid$ and T.E $\propto {\displaystyle \frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}}$; therefore

$$\equiv {\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\left({\displaystyle \frac{{\mathrm{z}}_{\mathrm{H}}}{{\mathrm{z}}_{\mathrm{Li}}}}\right)}^2\times {\left({\displaystyle \frac{{\mathrm{n}}_{\mathrm{Li}}}{{\mathrm{n}}_{\mathrm{H}}}}\right)}^2={\left({\displaystyle \frac{1}{3}}\right)}^2\times {\left({\displaystyle \frac{3}{2}}\right)}^2={\displaystyle \frac{1}{4}}$$

We thus have $({\displaystyle \frac{{\ell }_{\mathrm{H}}}{{\ell }_{\mathrm{Li}}}}={\displaystyle \frac{2}{3}}$ and ${\displaystyle \frac{{\mathrm{K}}_{\mathrm{H}}}{{\mathrm{K}}_{\mathrm{Li}}}}={\displaystyle \frac{1}{4}})$

Hence option (3) is correct.

Average

Bohr’s Model of the Atom

6. A hydrogen atom emits a photon of wavelength $\lambda$, in returning from the first excited state to ground state. Another hydrogen like ion of an atomic number $\mathrm{z}$, emits a photon of wavelength $(\alpha \lambda )$; in returning from its second excited state to its first excited state. The value of $\mathrm{z}$, then equals

(1) $\sqrt{{\displaystyle \frac{3}{5\alpha }}}$

(2) $\sqrt{{\displaystyle \frac{9}{2\alpha }}}$

(3) $\sqrt{{\displaystyle \frac{27}{5\alpha }}}$

(4) $\sqrt{{\displaystyle \frac{5}{27\alpha }}}$

Show Answer

Correct answer: (3)

Solution:

For hydrogen atom, the given transition, corresponds to an electron moving from ${\mathrm{n}}_2=2$, to ${\mathrm{n}}_1=1$. Therefore,

${\displaystyle \frac{1}{\lambda }}=\mathrm{R}[{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}}]={\displaystyle \frac{3\mathrm{R}}{4}}$ $$ …….(1)

For the given ion; the electron jumps from $n_2=3$, to $n_1=2$. Therefore,

${\displaystyle \frac{1}{\alpha \lambda }}={\mathrm{Rz}}^2({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}})={\displaystyle \frac{5{\mathrm{Rz}}^2}{36}}$ $$ …….(2)

From equations (1) and (2), we have ${\displaystyle \frac{3\mathrm{R}}{4\alpha }}={\displaystyle \frac{5{\mathrm{Rz}}^2}{36}}$

$\therefore \mathrm{z}=\sqrt{{\displaystyle \frac{27}{5\alpha }}}$

Hence option (3) is correct.

Average

Bohr’s Model of the Atom

7. A diatomic molecules has a moment of inertia I. By Bohr’s quantisation condition, its rotational energy in the ${\mathrm{n}}^{\text{th }}$ level $(\mathrm{n}\ne 0)$ is

(1) ${\displaystyle \frac{{\mathrm{h}}^2}{8{\pi }^2{\mathrm{In}}^2}}$

(2) ${\displaystyle \frac{{\mathrm{h}}^2}{8{\pi }^2\mathrm{In}}}$

(3) ${\displaystyle \frac{\mathrm{n}{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}}$

(4) ${\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}}$

Show Answer

Correct answer: (4)

Solution:

Let ${\omega }_{\mathrm{n}}$ be the angular speed of molecule in ${\mathrm{n}}^{\text{th }}$ allowed state. Its angular momentum is $L_{\mathrm{n}}=I{\omega }_{\mathrm{n}}$. From Bohr’s postulate, we have

$${\mathrm{L}}_{\mathrm{n}}(=\mathrm{I}{\omega }_{\mathrm{n}})={\displaystyle \frac{\mathrm{nh}}{2\pi }}\therefore {\omega }_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\pi \mathrm{I}}}$$

The rotational energy of the molecule $={\displaystyle \frac{1}{2}}\mathrm{I}{\omega }_{\mathrm{n}}^2={\displaystyle \frac{1}{2}}\mathrm{I}{\left({\displaystyle \frac{\mathrm{nh}}{2\pi \mathrm{I}}}\right)}^2={\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}}$

Hence option (4) is correct.

Difficult

Using Bohr’s Postulates

8. A molecule of, $\mathrm{CO}$, a diatomic molecule, is made up of a carbon atom (mass $=12\mathrm{u})$ and an oxygen atom (mass $=16u)$, separated by a distance $r$. It is assumed that the rotational K.E, of the molecule, about its centre of mass, is quantized as per Bohr’s postulates for the hydrogen atom. If the frequency of the photon, that can excite the molecule from its (rotational kinetic energy) ground state to its first excited state is $(4/\pi \times {10}^{11})\mathbf{H}\mathbf{z}$ (and the Planck’s constant equals $2\pi \times {10}^{-34}\mathrm{J}-\mathrm{s}$ ), the distance $\mathrm{r}$, and the moment of inertia, I, of the molecule, about its centre of mass, are (nearly) equal, respectively, to

(1) $1.28\times {10}^{-10}\mathrm{m}$ and $1.88\times {10}^{-46}\mathrm{kg}-{\mathrm{m}}^2$

(2) $2.4\times {10}^{-10}\mathrm{m}$ and $2.76\times {10}^{-46}\mathrm{kg}-{\mathrm{m}}^2$

(3) $4.4\times {10}^{-11}\mathrm{m}$ and $4.67\times {10}^{-47}\mathrm{kg}-{\mathrm{m}}^2$

(4) $1.9\times {10}^{-10}\mathrm{m}$ and $1.17\times {10}^{-47}\mathrm{kg}-{\mathrm{m}}^2$

Show Answer

Correct answer: (1)

Solution:

Let $\omega$ be the angular frequency of rotation. The angular momentum $\mathrm{L}(=\mathrm{I}\omega )$ is then given, as per Bohr’s postulate, by ${\mathrm{L}}_{\mathrm{n}}=\mathrm{I}{\omega }_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\pi }}$, hence ${\omega }_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\pi \mathrm{I}}}$

The rotational, K.E., (K), equals ${\displaystyle \frac{1}{2}}\mathrm{I}{\omega }^2$. Hence the rotational kinetic energy levels are given by

$${\mathrm{K}}_{\mathrm{n}}={\displaystyle \frac{1}{2}}\mathrm{I}{\omega }_{\mathrm{n}}^2={\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}}$$

We, therefore, have

$$\begin{array}{rl}& ({\displaystyle \frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}}-{\displaystyle \frac{{\mathrm{I}}^2{\mathrm{h}}^2}{8{\pi }^2\mathrm{I}}})=\mathrm{h}({\displaystyle \frac{4}{\pi }}\times {10}^{11})\\ & \therefore {\displaystyle \frac{8\pi \mathrm{I}}{3\mathrm{h}}}={\displaystyle \frac{10}{4}}\times {10}^{-11}\\ & \therefore \mathrm{I}=\left({\displaystyle \frac{3\times 2\pi \times {10}^{-34}\times {10}^{-11}}{4\times 8\pi }}\right)\mathrm{kg}-{\mathrm{m}}^2=1.88\times {10}^{-46}\mathrm{kg}-{\mathrm{m}}^2\\ & \text{ Now }\mathrm{I}={\mathrm{m}}_1{\mathrm{r}}_1^2+{\mathrm{m}}_2{\mathrm{r}}_2^2=\left({\displaystyle \frac{{\mathrm{m}}_1{\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}}\right){\mathrm{r}}^2\\ & \therefore \mathrm{R}={\left({\displaystyle \frac{28\times 1.88\times {10}^{-46}}{192\times 1.67\times {10}^{-27}}}\right)}^{1/2}\mathrm{m}=1.28\times {10}^{-10}\mathrm{m}\end{array}$$

Hence option (1) is correct

Difficult

Bohr’s Model of the Atom

9. A hydrogen atom, at rest, emits the $\beta$-line of its Balmer series. The recoil energy of the hydrogen atom, in terms of $R$ (= Rydbreg’s constant for hydrogen), $h$ (=Planck’s constant) and M (= Mass of hydrogen atom), would be

(1) ${\displaystyle \frac{9{\mathrm{h}}^2{\mathrm{R}}^2}{256\mathrm{M}}}$

(2) ${\displaystyle \frac{9{\mathrm{h}}^2{\mathrm{R}}^2}{512\mathrm{M}}}$

(3) ${\displaystyle \frac{25{\mathrm{h}}^2{\mathrm{R}}^2}{1296\mathrm{M}}}$

(4) ${\displaystyle \frac{25{\mathrm{h}}^2{\mathrm{R}}^2}{512\mathrm{M}}}$

Correct answer: (2)

Solution:

The $\beta$-line, in the Balmer series, corresponds to a transition from ${\mathrm{n}}_2=4$ to ${\mathrm{n}}_1=2$. The wavelength, $\lambda$, of this $\beta$-line, is given by

$${\displaystyle \frac{1}{\lambda }}=\mathrm{R}[{\displaystyle \frac{1}{(2{)}^2}}-{\displaystyle \frac{1}{(4{)}^2}}]={\displaystyle \frac{3\mathrm{R}}{16}}$$

The linear momentum of the emitted photon $=\mathrm{p}={\displaystyle \frac{\mathrm{h}}{\lambda }}$ or $\mathrm{p}={\displaystyle \frac{3\mathrm{hR}}{16}}$

The hydrogen atom is initially at rest. To conserve linear momentum, the recoil momentum of the hydrogen atom, is also p, directed in a direction opposite to the direction of motion of the photon. Hence the recoil

energy of the hydrogen atom $={\displaystyle \frac{{\mathrm{p}}^2}{2M}}={\displaystyle \frac{9{\mathrm{h}}^2{\mathrm{R}}^2}{512\mathrm{M}}}$

Hence option (2) is correct.

Difficult

Bohr’s Model of the Atom

10. A hydrogen like atom has the energy $\left({\mathrm{E}}_{\mathrm{n}}\right)$ in its nth allowed orbit, given by ${\mathrm{E}}_{\mathrm{n}}=\left({\displaystyle \frac{\alpha }{n^2}}\right)$ where $\alpha$ is a constant. The atom is in its ${\mathrm{m}}^{\text{th }}$ excited state. It emits rediations in coming to ground state directly from the excited state. The emitted radiations are incident on a photosensitive material of threshold wavelength ${\lambda }_0$. The maximum K.E, of the emitted photoelectrons, equals twice the photoelectric work function of the material. The quantum number $\mathrm{m}$ equals

(1) ${\left({\displaystyle \frac{{\lambda }_0\alpha }{{\lambda }_0\alpha -2\mathrm{hc}}}\right)}^{1/2}+1$

(2) ${\left({\displaystyle \frac{{\lambda }_0\alpha }{{\lambda }_0\alpha -2\mathrm{hc}}}\right)}^{1/2}-1$

(3) ${\left({\displaystyle \frac{{\lambda }_0\alpha }{{\lambda }_0\alpha -\mathrm{hc}}}\right)}^{1/2}+1$

(4) ${\left({\displaystyle \frac{{\lambda }_0\alpha }{{\lambda }_0\alpha -3\mathrm{hc}}}\right)}^{1/2}-1$

Show Answer

Correct answer: (4)

Solution:

We have ${\mathrm{w}}_0=$ work function of photosensitive material $={\displaystyle \frac{\mathrm{hc}}{{\lambda }_0}}$. Let $\mathrm{E}=\mathrm{h}v$, be the energy of photon incident on the photosensitive material. FromEinstein’s equation of photoelectric effect, we have

$2{\mathrm{w}}_0=\mathrm{E}-{\mathrm{w}}_0\therefore \mathrm{E}=3{\mathrm{w}}_0={\displaystyle \frac{3\mathrm{hc}}{{\lambda }_0}}$ $$ …….(1)

From Bohr’s theory, $\mathrm{E}=({\mathrm{E}}_{\mathrm{m}+1}-{\mathrm{E}}_1)=(\alpha )[1-{\displaystyle \frac{1}{(\mathrm{m}+1{)}^2}}]$ $$ …….(2)

From Eqn. (1) and (2) we have,

$$\begin{array}{rl}& {\displaystyle \frac{3\mathrm{hc}}{{\lambda }_0}}=\alpha [1-{\displaystyle \frac{1}{(\mathrm{m}+1{)}^2}}]\text{ or }{\displaystyle \frac{1}{(\mathrm{m}+1{)}^2}}=[1-{\displaystyle \frac{3\mathrm{hc}}{\alpha {\lambda }_0}}]\\ \therefore \mathrm{m}& ={\left[{\displaystyle \frac{\alpha {\lambda }_0}{\alpha {\lambda }_0-3\mathrm{hc}}}\right]}^{1/2}-1\end{array}$$

Hence option (4) is correct.

Difficult

Bohr’s Model of the Atom

11. A particle is restricted to move along the $x$-axis between $x=0$ and $x=a$ where $\mathrm{x}$ is a given distance, of the order of a nanometer. The allowed energies of the particle, correspond to formation of stationary waves, with nodes at $x=0$ and $x=a$. The wavelength, of the standing waves, is related to the linear momentum, $p$; of the particle, by the deBroglie relation. The energy, $E$, of the particle, is given by; $E=p^2/2m$. The quantised energy values $\left(E_n\right)$, of the particle, $($ where $n=$ quantum number; $(n=1,2,3,\dots \dots$.$))$ and its velocity $\left(v_n\right)$, then vary, respectively, as

(1) ${\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{a}}^{-2};{\mathrm{v}}_{\mathrm{n}}\propto \mathrm{n}$

(2) ${\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{a}}^{-3/2};v\propto {\mathrm{n}}^{-3/2}$

(3) ${\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{a}}^{-1};\mathrm{v}\propto {\mathrm{n}}^{-1}$

(4) ${\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{a}}^2;\mathrm{v}\propto \mathrm{n}$

Show Answer

Correct answer: (1)

Solution:

Figure shows the alternate nodes and antinodes in the region between $x=0$ and $x=\mathrm{a}$. In the ${\mathrm{n}}^{\text{th }}$ mode, there are n nodal points. The wavelength, ${\lambda }_n$, of the stationary waves, formed is, therefore, given by

$$\mathrm{n}\left({\displaystyle \frac{{\lambda }_{\mathrm{n}}}{2}}\right)=\mathrm{a};\therefore {\lambda }_{\mathrm{n}}={\displaystyle \frac{2\mathrm{a}}{\mathrm{n}}}$$

Let ${\mathrm{p}}_{\mathrm{n}}$ be the linear momentum of the particle. We are given that ${\lambda }_{\mathrm{n}}={\displaystyle \frac{\mathrm{h}}{{\mathrm{p}}_{\mathrm{n}}}}={\displaystyle \frac{\mathrm{h}}{\sqrt{2{\mathrm{mE}}_{\mathrm{n}}}}}$

$\therefore {\displaystyle \frac{2\mathrm{a}}{\mathrm{n}}}={\displaystyle \frac{\mathrm{h}}{\sqrt{2{\mathrm{mE}}_{\mathrm{n}}}}}$

$\therefore E_n={\displaystyle \frac{n^2{h}^2}{8m^2}}$

For a given value of m; $E_n\propto a^{-2}$

Also ${\mathrm{mv}}_{\mathrm{n}}\propto {\mathrm{p}}_{\mathrm{n}}={\displaystyle \frac{\mathrm{hn}}{2\mathrm{a}}}\therefore {\mathrm{v}}_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\mathrm{ma}}}$

$\therefore {\mathrm{v}}_{\mathrm{n}}\propto \mathrm{n}$

Hence option(1) is correct.

Difficult

Bohr’s Model of the Atom

12. The radius of an orbit of the electron, in a hydrogen like atom, is $4.5a_0$, where $a_0$ is the Bohr’s radius for hydrogen. Its corresponding orbital angular momentum is ${\displaystyle \frac{3\mathrm{h}}{2\pi }}$. (h=Planck’s constant; $R=$ Rydberg’s constant). The wave length, this atom would not be emitting, as it de-exites itself, is

(1) ${\displaystyle \frac{9}{32\mathrm{R}}}$

(2) ${\displaystyle \frac{9}{16\mathrm{R}}}$

(3) ${\displaystyle \frac{9}{5\mathrm{R}}}$

(4) ${\displaystyle \frac{1}{3\mathrm{R}}}$

Show Answer

Correct answer: (2)

Solution:

Since we are given that angular momentum $={\displaystyle \frac{3\mathrm{h}}{2\pi }}=\mathrm{n}{\displaystyle \frac{\mathrm{h}}{2\pi }}$, hence $\mathrm{n}=3$

The radius, ${\mathrm{r}}_{\mathrm{n}}$, of the ${\mathrm{n}}^{\text{th }}$ allowed orbit, is given by ${\mathrm{r}}_{\mathrm{n}}\propto {\displaystyle \frac{{\mathrm{n}}^2}{\mathrm{z}}}$

$\therefore 4.5={\displaystyle \frac{(3{)}^2}{\mathrm{z}}}\therefore \mathrm{z}=2$

The wavelength, $\lambda$, of the emitted radiation, is

$${\displaystyle \frac{1}{\lambda }}={\mathrm{Rz}}^2[{\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}}]$$

The three different ways, the atom can de-excute itself(from the state with $n=3$ ) are

(i) ${\mathrm{n}}_2=3\to {\mathrm{n}}_1=1$; (ii) ${\mathrm{n}}_2=3\to {\mathrm{n}}_1=2$; and (iii) ${\mathrm{n}}_2=2\to {\mathrm{n}}_1=1$.

The corresponding wavelengths are

$$\begin{array}{rlrl}{\displaystyle \frac{1}{{\lambda }_1}}& =(2{)}^2\mathrm{R}[{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{9}}]={\displaystyle \frac{32\mathrm{R}}{9}}& & \therefore {\lambda }_1={\displaystyle \frac{9}{32\mathrm{R}}}\\ {\displaystyle \frac{1}{{\lambda }_2}}& =(2{)}^2\mathrm{R}[{\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}}]={\displaystyle \frac{5\mathrm{R}}{9}}& & \therefore {\lambda }_2={\displaystyle \frac{9}{5\mathrm{R}}}\\ \text{ and }{\displaystyle \frac{1}{{\lambda }_3}}& =(2{)}^2\mathrm{R}[{\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}}]=3\mathrm{R}& & \therefore {\lambda }_3={\displaystyle \frac{1}{3\mathrm{R}}}\end{array}$$

It follows that, out of the given values $\lambda ={\displaystyle \frac{9}{16\mathrm{R}}}$, is the value, that the atom would not emit.

Hence option (2) is correct.

Difficult

Bohr’s Model of the Atom

13. A hydrogen like atom, of atomic number $\mathrm{z}$ is in its $(2\mathrm{n}{)}^{\text{th }}$ state. From this state, the atom can emit a photon of maximum energy $204\mathrm{eV}$. It makes a transction, from this excited state, to a state of quantum number $n$, emitting a photon of energy $40.8\mathrm{eV}$. For the atom

(1) $\mathrm{z}=4;\mathrm{n}=2$

(2) $\mathrm{z}=3;\mathrm{n}=3$

(3) $\mathrm{z}=4;\mathrm{n}=4$

(4) $\mathrm{z}=3;\mathrm{n}=4$

Show Answer

Correct answer: (1)

Solution:

Let ${\mathrm{E}}_1$ be the energy of the atom in its ground state. For atom in $(2\mathrm{n}{)}^{\mathrm{th}}$ state ${\mathrm{E}}_{2\mathrm{n}}=\left[{\displaystyle \frac{{\mathrm{E}}_1}{(2\mathrm{n}{)}^2}}\right]$. The maximum energy photon is emitted when electron jumps from this state to the ground state, i.e.,

$({\displaystyle \frac{E_1}{4{\mathrm{n}}^2}}-E_1)=204\mathrm{eV}$

$\therefore E_1({\displaystyle \frac{1}{4{\mathrm{n}}^2}}-1)=204$ $$ …….(1)

When electron jumps, from ${\mathrm{n}}_2(=2\mathrm{n})$ to ${\mathrm{n}}_1(=\mathrm{n})$ state, the energy of emitted photon is $40.8\mathrm{eV}$. Therefore

${\mathrm{E}}_1[{\displaystyle \frac{1}{4{\mathrm{n}}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}]=40.8\text{ or }={\displaystyle \frac{3{\mathrm{E}}_1}{4{\mathrm{n}}^2}}=40.8$ $$ …….(2)

From Eqns. (1) and (2), we have

$$\begin{array}{}\text{(3)}& {\displaystyle \frac{(1-{\displaystyle \frac{1}{4{\mathrm{n}}^2}})}{\left({\displaystyle \frac{3}{4{\mathrm{n}}^2}}\right)}}={\displaystyle \frac{204}{40.8}}=5\end{array}$$

$\therefore \mathrm{n}=2$

Substituting this value of $n$, in Eqn. (2), we get

$$\begin{array}{}\text{(4)}& E_n=-{\displaystyle \frac{4\times (2{)}^2\times 40.8}{3}}\mathrm{eV}=-217.6\mathrm{eV}\end{array}$$

But ${\mathrm{E}}_1={\mathrm{E}}_1$ (hydrogen) ${\mathrm{z}}^2=[-(13.6){\mathrm{z}}^2]\mathrm{eV}$

$\therefore {\mathrm{z}}^2={\displaystyle \frac{-217.6}{(-13.6)}}=16$

$\therefore \mathrm{z}=4$

Hence option(1) is correct.

Difficult

Bohr’s Model of the Atom

14. A mixture, of hydrogen $(\mathrm{H})$ and singly ionised helium $\left({\mathrm{He}}^{+}\right)$, has the $\mathrm{H}$ and ${\mathrm{He}}^{+}$ions excited to their respective first excited states. Due to collosions, $H$ atoms transfer their entire excitation energy to the ${\mathrm{He}}^{+}$ion. Applying Bohr’s model, the quantum number, $\mathrm{n}$, of the finally populated state of ${\mathrm{He}}^{+}$ions, is

(1) 2

(2) 3

(3) 4

(4) 5

Show Answer

Correct answer: (3)

Solution:

For hydrogen atom; $E_n={\displaystyle \frac{-13.6}{n^2}}\mathrm{eV}$

Hence ${\mathrm{E}}_2={\displaystyle \frac{-13.6}{4}}\mathrm{eV}=-3.4\mathrm{eV}$

Excitation energy gained by the $\mathrm{H}$-atom, in the first excited state (i.e. $\mathrm{n}=2$ ), is

$${\mathrm{E}}_2-{\mathrm{E}}_1=(13.6-3.4)\mathrm{eV}=10.2\mathrm{eV}$$

For ${\mathrm{He}}^{+}$ion, $(\mathrm{z}=2)$; therefore, its energy, in an, excited state $(\mathrm{n}=\mathrm{n})$, is

$${\mathrm{E}}_{\mathrm{n}}^{\prime }-{\displaystyle \frac{(13.6)(2{)}^2}{{\mathrm{n}}^2}}=-{\displaystyle \frac{54.4}{{\mathrm{n}}^2}}\mathrm{eV}$$

$\therefore$ The energy, of ${\mathrm{He}}^{+}$ion, is the first excited state (i.e., $\mathrm{n}=2$ ), is ${\mathrm{E}}_2^{\prime }=-13.6\mathrm{eV}$

The energy $\mathrm{E}$ “, of the ${\mathrm{He}}^{+}$ion, after collosion with $\mathrm{H}$-atoms;

$${\mathrm{E}}^{\prime \prime }-{\mathrm{E}}_2^{\prime }+10.2\mathrm{eV}=-3.4\mathrm{eV}$$

Let $\mathrm{n}$ be the quantum number of state, corresponding to an energy of $-3.4\mathrm{eV}$, in the ${\mathrm{He}}^{+}$ion. Then

$$\begin{array}{rl}& -3.4=-{\displaystyle \frac{54.4}{{\mathrm{n}}^2}}\\ & \therefore {\mathrm{n}}^2={\displaystyle \frac{54.4}{3.4}}=16\\ & \therefore \mathrm{n}=4\end{array}$$

Hence option (3) is correct.

Average

Bohr’s Model of the Atom

15. A single electron revolves around a stationary nucleus of charge ze; ( $e$ = magnitude of charge on electron). A photon, of energy $47.2\mathrm{eV}$, can excite this atom from its first excited state to its second excited state. The kinetic energy, $K$; the potential energy $U$, and the total energy, $\mathrm{E}$, of this atom, in its ground state, are:

(1) $\mathrm{K}=340\mathrm{eV},\mathrm{U}=-170\mathrm{eV},\mathrm{E}=-340\mathrm{eV}$

(2) $\mathrm{K}=340\mathrm{eV},\mathrm{U}=-680\mathrm{eV},\mathrm{E}=-340\mathrm{eV}$

(3) $\mathrm{K}=85\mathrm{eV},\mathrm{U}=-170\mathrm{eV},\mathrm{E}=-85\mathrm{eV}$

(4) $\mathrm{K}=170\mathrm{eV},\mathrm{U}=-340\mathrm{eV},\mathrm{E}=-170\mathrm{eV}$

Show Answer

Correct answer: (2)

Solution:

The energy, ${\mathrm{E}}_{\mathrm{n}}$, of this atom, in its ${\mathrm{n}}^{\text{th }}$ allowed state, is ${\mathrm{E}}_{\mathrm{n}}=-{\displaystyle \frac{(13.6){\mathrm{z}}^2}{{\mathrm{n}}^2}}\mathrm{eV}$

When this atom goes, from its first excited to its second excited state; $n_1=2$ and $n_2=3$. We are given that

$$\begin{array}{c}{\mathrm{E}}_3-{\mathrm{E}}_2=47.2\mathrm{eV}\\ \therefore \left({\mathrm{z}}^2\right)(13.6)[{\displaystyle \frac{1}{(2{)}^2}}-{\displaystyle \frac{1}{(3{)}^2}}]=47.2\end{array}$$

$\therefore {\mathrm{z}}^2={\displaystyle \frac{47.2\times 36}{5\times 13.6}}=25$

$\therefore \mathrm{z}=5$

$\therefore$ For the given atom,

$$E_n=-{\displaystyle \frac{13.6(5{)}^2}{n^2}}eV=-{\displaystyle \frac{340}{n^2}}eV$$

For the ground state of atom, $\mathrm{n}=1$. Therefore,

$\mathrm{E}=-340\mathrm{eV};\mathrm{K}=|\mathrm{E}|=340\mathrm{eV}$

and

$$\mathrm{U}=2\mathrm{E}=-680\mathrm{eV}$$

Hence option(2) is correct.

Difficult

Bohr’s Model of the Atom

16. In a hypothetical atom, the potential energy $(U)$, of an electron and a proton, distance $r$ apart, is given by $U=-{\displaystyle \frac{k^2}{3r^3}}$ where $k$ is a constant. Applying Bohr’s theory to this atom, the speed $v_n$, and the total energy, $E_n$, of the electron, in its $n^{\text{th }}$ allowed orbit, would depend on ’ $n$ ‘, as:

(1) ${\mathrm{V}}_{\mathrm{n}}\propto {\mathrm{n}}^3;{\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{n}}^6$

(2) ${\mathrm{v}}_{\mathrm{n}}\propto {\mathrm{n}}^{5/2};{\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{n}}^5$

(3) ${\mathrm{v}}_{\mathrm{n}}\propto {\mathrm{n}}^2;{\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{n}}^4$

(4) ${\mathrm{V}}_{\mathrm{n}}\propto {\mathrm{n}}^{-3};{\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{n}}^{-6}$

Show Answer

Correct answer: (1)

Solution:

Given $\mathrm{U}=-{\displaystyle \frac{{\mathrm{ke}}^2}{3{\mathrm{r}}^3}}$. The force, $\mathrm{F}$, between the electron and the proton, distant $\mathrm{r}$ apart, is

$\mathrm{F}=-{\displaystyle \frac{\mathrm{dU}}{\mathrm{dr}}}={\displaystyle \frac{{\mathrm{ke}}^2}{{\mathrm{r}}^4}}$ $$ …….(1)

For an electron, in its circular orbit of radius $r_n$, the speed $v_n$ (is the $n^{\text{th }}$ allowed orbit), would be given by

${\displaystyle \frac{m{v}_n^2}{r_n}}=F={\displaystyle \frac{{\mathrm{ke}}^2}{{\mathrm{r}}^4}}$ $$ …….(2)

From Bohr’s first postulate ${\mathrm{mv}}_{\mathrm{n}}{\mathrm{r}}_{\mathrm{n}}={\displaystyle \frac{\mathrm{nh}}{2\pi }}$ $$ …….(3)

From Eqn. (2) and (3), we get $v_n={\displaystyle \frac{n^3{h}^3}{8{\pi }^3{\mathrm{km}}^2{e}^2}}$

i.e. ${\mathrm{v}}_{\mathrm{n}}\propto {\mathrm{n}}^3$

The total energy, ${\mathrm{E}}_{\mathrm{n}}$, in the ${\mathrm{n}}^{\text{th }}$ allowed orbit, is ${\mathrm{E}}_{\mathrm{n}}=-{\mathrm{K}}_{\mathrm{n}}=-{\displaystyle \frac{1}{2}}{\mathrm{mv}}_{\mathrm{n}}^2$

$\therefore {\mathrm{E}}_{\mathrm{n}}\propto {\mathrm{n}}^6$

Hence option(1) is correct.

Average

Radioactivity

17. The (radio) activities, of two materials $X$ and $Y$, when they are having $N_1$ and $N_2$ atoms, are $R_1$ and $R_2$, respectively. It is observed that $R_2=3R_1$, when $N_1=2N_2$. If the half life of $X$ is 1 month, that of $Y$ would be

(1) 6 months

(2) 3 months

(3) $1/3$ months

(4) $1/6$ months

Show Answer

Correct answer: (4)

Solution:

Let ${\lambda }_1$ and ${\lambda }_2$ and be the disintegration constants of $X$ and $Y.N_1$ and $N_2$ are their number of atoms, at the given instant. Then

${\mathrm{R}}_1={\lambda }_1{\mathrm{N}}_1\text{ and }{\mathrm{R}}_2={\lambda }_2{\mathrm{N}}_2$

$\therefore {\displaystyle \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}}=\left({\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}\right)\left({\displaystyle \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}}\right)$

$\therefore {\displaystyle \frac{1}{3}}=\left({\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}\right)\times 2$

$\therefore {\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}={\displaystyle \frac{1}{6}}$ $$ ……(1)

Let ${\tau }_1$ and ${\tau }_2$ be the half-life of $\mathrm{X}$ and $\mathrm{Y}$, respectively. We know that half-life is inversly proportional to the disintegration constant. Therefore

$${\displaystyle \frac{{\tau }_2}{{\tau }_1}}={\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}={\displaystyle \frac{1}{6}}$$

Given ${\tau }_1=1$ month. Therefore, ${\tau }_2={\displaystyle \frac{1}{6}}$ month

Hence option(4) is correct.

Average

Radioactivity

18. Two radioactive materials $x_1$ and $x_2$, have decay constants $10\lambda$ and $\lambda$, respectively. It initially they have the same number of nuclie then the ratio of number of nuclei of $x_1$ to that of $x_2$ would equal $(1/e)$ after a time.

(1) ${\displaystyle \frac{1}{10\lambda }}$

(2) ${\displaystyle \frac{1}{11\lambda }}$

(3) ${\displaystyle \frac{11}{10\lambda }}$

(4) ${\displaystyle \frac{1}{9\lambda }}$

Show Answer

Correct answer: (4)

Solution:

We assume that the given condition gets satisfied at $\mathrm{t}=\mathrm{t}$.

Let ${\mathrm{N}}_0$ be the number of nuclei of $x_1$ and $x_2$ at $\mathrm{t}=0$. Let the number of nuclei of $x_1$ and $x_2$, present at $\mathrm{t}=\mathrm{t}$, be ${\mathrm{N}}_1$ and ${\mathrm{N}}_2$, respectively. From the law of radioactive disintegration, we have

$${\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{-10\lambda t}\text{ and }{\mathrm{N}}_2={\mathrm{N}}_0{\mathrm{e}}^{\lambda \mathrm{t}}$$

$\therefore {\displaystyle \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}}={\mathrm{e}}^{-9\lambda \mathrm{T}}={\displaystyle \frac{1}{\mathrm{e}}}={\mathrm{e}}^{-1}$

$\therefore 9\lambda t=1$

$\therefore \mathrm{t}={\displaystyle \frac{1}{9\lambda }}$

Hence option (4) is correct

Average

Radioactivity

19. A radioactive material, having a mean life of ${10}^9\mathrm{s}$, has an activity of ${10}^{10}$ disintegrations per second, at a given instant. If the mass of one atom of this material is ${10}^{-25}\mathrm{kg}$, the total mass (in $\mathrm{mg}$ ) of this material, present at the given instant, is

(1) One

(2) Half

(3) Two

(4) One and a half

Show Answer

Correct answer: (1)

Solution:

Let $\mathrm{N}$ be the number of radioactive nuclei present in the sample, at the given instant and let $\lambda$ be its disintegration constant. Then

Activity $=\left|{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}\right|=\lambda \mathrm{N}$

Since $\tau ={\displaystyle \frac{1}{\lambda }}$ we have activity $=\left({\displaystyle \frac{N}{\tau }}\right)$

Substituting given values, we get ${10}^{10}={\displaystyle \frac{\mathrm{N}}{\left({10}^9\right)}}$

$\therefore \mathrm{N}={10}^{19}$ atoms

$\therefore$ Mass of sample $=\mathrm{N}\times$ mass of one atom

$={10}^{19}\times {10}^{-25}\mathrm{kg}={10}^{-6}\mathrm{kg}=1\mathrm{mg}$

Hence option (1) is correct.

Average

Radioactivity

20. A radioactive nucleus has a half-life of one day. Starting with $N_0$ atoms, at $t=0;N_1$ is number of radioactive atoms present at $t_1=\mathbf{2}\mathbf{.}\mathbf{8}\mathbf{8}\mathbf{6}$ days and $N_2$ the is number of atoms present at $t_2=\mathbf{4}\mathbf{.}\mathbf{3}\mathbf{2}\mathbf{9}$ days. The number of atoms decaying, in the time interval $(t_2-t_1)$, expressed as a fraction of ${\mathrm{N}}_0$, is

(1) $\left({\displaystyle \frac{\mathrm{e}-1}{{\mathrm{e}}^2}}\right)$

(2) $\left({\displaystyle \frac{-\mathrm{e}+1}{{\mathrm{e}}^3}}\right)$

(3) $\left({\displaystyle \frac{{\mathrm{e}}^2-1}{{\mathrm{e}}^2+1}}\right)$

(4) $\left({\displaystyle \frac{{\mathrm{e}}^2-1}{{\mathrm{e}}^3}}\right)$

Show Answer

Correct answer: (2)

Solution:

Given, ${\tau }_{1/2}=$ Half-life $=1$ day $={\displaystyle \frac{0.693}{\lambda }}$ where $\lambda$ is the desintegration constant.

$\therefore \lambda =0.693$ day ${}^{-1}$

$\mathrm{N}$, the number of radioactive atoms present at $\mathrm{t}=\mathrm{t}$, is given by $\mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-0.693\times \mathrm{t}}$

At $\mathrm{t}={\mathrm{t}}_1=2.886$ days

$${\mathrm{N}}_1={\mathrm{N}}_0{\mathrm{e}}^{-0.693\times 2.886}\simeq {\mathrm{N}}_0{\mathrm{e}}^{-2}$$

$${\mathrm{N}}_2={\mathrm{N}}_0{\mathrm{e}}^{-0.693\times 4.329}={\mathrm{N}}_0{\mathrm{e}}^{-3}$$

The number of atom, $\mathrm{\Delta }\mathrm{N}$, that have disintegrated in time interval $({\mathrm{t}}_2-{\mathrm{t}}_1)$, is

$$\begin{array}{rl}\mathrm{\Delta }\mathrm{N}& =({\mathrm{N}}_2-{\mathrm{N}}_1)={\mathrm{N}}_0({\mathrm{e}}^{-3}-{\mathrm{e}}^{-2})\\ \therefore {\displaystyle \frac{\mathrm{\Delta }\mathrm{N}}{{\mathrm{N}}_0}}& ={\mathrm{e}}^{-3}-{\mathrm{e}}^{-2}=\left({\displaystyle \frac{1-\mathrm{e}}{{\mathrm{e}}^3}}\right)=\left({\displaystyle \frac{-\mathrm{e}+1}{{\mathrm{e}}^3}}\right)\end{array}$$

Hence option (2) is correct.

Difficult

Radioactivity

21. A radioactive element decays by $\beta$-emission. A $\beta$-counter shows a count of $(X)$ in a time equal to one mean life of the element. The same counter shows $Y$ counts in a time equal to three times the mean life. The percentage change, in the number of counts, in the time interval $(\tau )$ to $(3\tau )(\tau =$ mean life $)$, is

(1) $\left[{\displaystyle \frac{\mathrm{e}-1}{{\mathrm{e}}^2}}\right]\times 100$

(2) $\left[{\displaystyle \frac{\mathrm{e}+1}{{\mathrm{e}}^2}}\right]\times 100$

(3) $\left({\displaystyle \frac{{\mathrm{e}}^2-1}{{\mathrm{e}}^2+1}}\right)\times 100$

(4) $\left({\displaystyle \frac{\mathrm{e}-1}{2\mathrm{e}-1}}\right)\times 100$

Show Answer

Correct answer: (2)

Solution:

The total number of counts at any time is the number of nuclei that have disintegrated in that time. Therefore, if $\tau$ is the mean life of the element, we have

$$\mathrm{X}={\mathrm{N}}_0[1-{\mathrm{e}}^{-\lambda \tau }]={\mathrm{N}}_0[1-{\mathrm{e}}^{-1}](\because \tau ={\displaystyle \frac{1}{\lambda }},\text{ where }\lambda =\text{ decay constant })$$

and $\mathrm{Y}={\mathrm{N}}_0[1-{\mathrm{e}}^{-3\lambda \overline{\tau }}]={\mathrm{N}}_0[1-{\mathrm{e}}^{-3}]$

Percentage change in counts in the time interval $({\mathrm{t}}_1=\tau )$ to $({\mathrm{t}}_2=3\tau )$, is $=\left({\displaystyle \frac{\mathrm{Y}-\mathrm{X}}{\mathrm{X}}}\right)\times 100\mathrm{\%}$

$$\begin{array}{rl}& =\left[{\displaystyle \frac{{\mathrm{e}}^{-1}-{\mathrm{e}}^{-3}}{1-{\mathrm{e}}^{-1}}}\right]\times 100\mathrm{\%}=\left[{\displaystyle \frac{{\mathrm{e}}^3({\mathrm{e}}^{-1}-{\mathrm{e}}^{-3})}{{\mathrm{e}}^3(1-{\mathrm{e}}^{-1})}}\right]\\ & ={\displaystyle \frac{({\mathrm{e}}^2-1)}{{\mathrm{e}}^2(\mathrm{e}-1)}}100\mathrm{\%}=\left({\displaystyle \frac{\mathrm{e}+1}{{\mathrm{e}}^2}}\right)\times 100\mathrm{\%}\end{array}$$

Hence option (2) is correct.

Average

Radioactivity

22. ${\mathbf{N}}_1$ is number of radioactive atoms of an element (of mean life 10 days), present at the end of 10 days. $N_2$ is the total number of atoms that have disintegrated in a time of 30 days. The $\mathrm{ratio}\left({\mathrm{N}}_1/{\mathrm{N}}_2\right)$, is

(1) ${\mathrm{e}}^2$

(2) $\left({\displaystyle \frac{{\mathrm{e}}^2}{{\mathrm{e}}^2-1}}\right)$

(3) $\left({\displaystyle \frac{{\mathrm{e}}^2}{{\mathrm{e}}^3-1}}\right)$

(4) $\left({\displaystyle \frac{{\mathrm{e}}^2-1}{{\mathrm{e}}^3-1}}\right)$

Show Answer

Correct answer: (3)

Solution:

Let ${\mathrm{N}}_0$ be number of atoms at $\mathrm{t}=0$

The disintegration constant $\lambda =1/10$ day ${}^{-1}$. Then if $(N=N_1)$, at $(t_1=\tau =10$ day $)$, we have

${\mathrm{N}}_1={\mathrm{N}}_0\left({\mathrm{e}}^{-1}\right)$ $$……(1)

At $t_2$ (= 30 days), the total number of atoms, that have disintegrated, is

${\mathrm{N}}_2={\mathrm{N}}_0[1-{\mathrm{e}}^{-3}]$ $$……(2)

$\therefore {\displaystyle \frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}}={\displaystyle \frac{{\mathrm{e}}^{-1}}{1-{\mathrm{e}}^{-3}}}={\displaystyle \frac{{\mathrm{e}}^2}{{\mathrm{e}}^3-1}}$

Hence option(3) is correct.

Average

Radioactivity

23. A radioactive nucleus ${}_{\mathrm{Z}}^A\mathrm{X}$ undergoes successive disintegrations, emitting $\mathit{x}\mathit{\alpha }$-particles followed by (positive) $\mathit{\beta }$-particles. The mass number $\left({\mathbf{A}}_1\right)$, and atomic number $\left({\mathrm{z}}_1\right)$, of the final daughter nucleus, are:

(1) ${\mathrm{A}}_1=(\mathrm{A}-2x);{\mathrm{z}}_1=(\mathrm{z}-2x+\mathrm{y})$

(2) ${\mathrm{A}}_1=(\mathrm{A}-4x);{\mathrm{z}}_1=(\mathrm{z}-2x-\mathrm{y})$

(3) ${\mathrm{A}}_1=(\mathrm{A}-4x);{\mathrm{z}}_1=(\mathrm{z}-2x+\mathrm{y})$

(4) ${\mathrm{A}}_1=(\mathrm{A}-2x);{\mathrm{z}}_1=(\mathrm{z}-x+\mathrm{y})$

Show Answer

Correct answer: (2)

Solution:

The emission of an $\alpha$-particle decreases mass number by 4 and atomic number by 2 . Therefore; after emission of $x,\alpha$-particles

$${\mathrm{A}}^{\prime }=(\mathrm{A}-4x),{\mathrm{z}}^{\prime }=(\mathrm{z}-2x)$$

The emission of a positive $\beta$-particle does not change the mass number but the atomic number decreases by one. Therefore, after emission of y positive $\beta$-particles, we have

$$\begin{array}{rl}& A_1=A^{\prime }=(A-4x)\\ & z_1=(z^{\prime }-y)=(z-2x-y)\end{array}$$

Hence option (2) is correct.

Difficult

Radioactivity

24. A radioactive nucleus (of mass number $A$ and atomic number $z$ ) has 84 protons and 136 neutrons. The nucleus undergoes successive disintegrations emittnig 3 ( $\alpha$ particles); 4

( $\beta$ particles) and 2 ( $\gamma$ ray photons). The percentage change, in the $(A/z)$ ratio, after all these disintegrations is (nearly)

(1) -3.05

(2) +1.91

(3) -0.38

(4) +7.25

Show Answer

Correct answer: (1)

Solution:

The mass number of given nucleus $=\mathrm{A}=$ number of protons + number of neutrons $=(84+136)=220$

The atomic number of the nucleus $=\mathrm{z}=$ number of protons $=84$

$\therefore$ Initial value of $=\left({\displaystyle \frac{\mathrm{A}}{\mathrm{z}}}\right)={\displaystyle \frac{220}{84}}=2.62$

We know that mass number decreases by 4 , and atomic number by 2 , when one $\alpha$-particle in emitted. The emission of a $\beta$-particle does not change the mass number but atomic number increases by one. There is no change in mass number, and atomic number, due to emission of a $\gamma$-ray photon. Let ${\mathrm{A}}_1$ and ${\mathrm{z}}_1$ be the mass number and atomic number, of the final nucleus, obtained after the given disintegrations. Then

$$\begin{array}{rl}& A_1=(220)-(4\times 3)=208\\ & z_1=[(84)-(4\times 3)]+(4\times 1)=82\end{array}$$

$\therefore$ Final value of ${\displaystyle \frac{{\mathrm{A}}_1}{{\mathrm{z}}_1}}={\displaystyle \frac{208}{82}}\simeq 2.54$

Percentage change in $\left({\displaystyle \frac{\mathrm{A}}{\mathrm{z}}}\right)$ ratio $=\left[{\displaystyle \frac{2.54-2.62}{2.62}}\right]\times 100\mathrm{\%}=-3.05\mathrm{\%}$

Hence option(1) is correct.

Difficult

Radioactivity

25. A radioactive nucleus, $X$, disintegrates into another nucleus $Y$, which in turn, disintegrates to form a nucles $z$. The disintegration constants, of $X$ and $Y$, are ${\lambda }_1$ and ${\lambda }_2$ respectively. At $t=0$; starting with $N_0$ atoms of $X$, the number of atoms of $X$ and $Y$, present at $t=t$, are $N_1$ and $N_2$, respectively. The ratio, of the instantaneous rates of disintegration, of $X$ and $Y$, is

(1) ${\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}$

(2) ${\displaystyle \frac{{\lambda }_1{N}_1}{{\lambda }_2{N}_2}}$

(3) ${({\displaystyle \frac{{\lambda }_2{\mathrm{N}}_2}{{\lambda }_1{\mathrm{N}}_1}}-1)}^{-1}$

(4) ${({\displaystyle \frac{{\lambda }_1{N}_1}{{\lambda }_2{N}_2}}+1)}^{-1}$

Show Answer

Correct answer: (3)

Solution:

${\mathrm{R}}_{\mathrm{X}}=$ The instantaneous rate of disintegration of $\mathrm{X}=-{\lambda }_1{\mathrm{N}}_1$ $$ …….(1)

When one atom of $X$ disintegrates, one atom of $Y$ is produced. $Att=t$; the rate of production $Y$, equals the rate of disintegration of $\mathrm{X}(={\lambda }_1{\mathrm{N}}_1)$. The instantaneous rate of decrease of $\mathrm{Y}$, due to its own decay into $\mathrm{Z}$, is $(-{\lambda }_2{N}_2)$. The net rate of change of the number of along of $Y$, is

${\mathrm{R}}_{\mathrm{Y}}={\displaystyle \frac{{\mathrm{dN}}_2}{\mathrm{dt}}}={\lambda }_1{\mathrm{N}}_1-{\lambda }_2{\mathrm{N}}_2$ $$ …….(2)

$\therefore {\displaystyle \frac{{\mathrm{R}}_{\mathrm{X}}}{{\mathrm{R}}_{\mathrm{Y}}}}={\displaystyle \frac{-{\lambda }_1{\mathrm{N}}_1}{{\lambda }_1{\mathrm{N}}_1-{\lambda }_2{\mathrm{N}}_2}}={\displaystyle \frac{1}{({\displaystyle \frac{{\lambda }_2{\mathrm{N}}_2}{{\lambda }_1{\mathrm{N}}_1}}-1)}}={({\displaystyle \frac{{\lambda }_2{\mathrm{N}}_2}{{\lambda }_1{\mathrm{N}}_1}}-1)}^{-1}$

Hence option(3) is correct.

Easy

Nucleus

26. A star can convert all its helium (He) nuclei completely into oxygen nuclei. The energy released, per oxygen nuclei formation, is (Mass of nucleus $\mathrm{He}=\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{0}\mathbf{2}\mathbf{6}\mathrm{u}$; mass of oxygen nucleus $=15.9994\mathrm{u})$

(1) $7.6\mathrm{meV}$

(2) $56.12\mathrm{meV}$

(3) $10.24\mathrm{meV}$

(4) $23.4\mathrm{meV}$

Show Answer

Correct answer: (3)

Solution:

The nuclear reaction is $4\left({}_2^4\mathrm{He}\right)\to {}_8^{16}\mathrm{O}$

$\therefore$ Mass defect $=\mathrm{\Delta }\mathrm{M}=[-15.9994+4\times (4.0026)]\mathrm{u}=0.011$

$\therefore$ Energy released (per oxygen nucleus) $=0.011\times 931\mathrm{meV}=10.24\mathrm{meV}$

Hence option (3) is correct.

Average

Nucleus

27. The only correct statement is:

(1) The nucleus ${}_3^7\mathrm{Li}$ can emit an $\alpha$-particle

(2) The nucleus of ${}_{82}^{206}\mathrm{Pb}$ can emit a proton

(3) Deutron and alpha particle can undergo complete fusion

(4) The nuclei ${}_{30}^{70}\mathrm{Zn}$ and ${}_{34}^{82}\mathrm{Sn}$ can undergo complete fusion

[Given; mass of $={}_1^1\mathrm{H}=1.007825\mathrm{u};{}_1^2\mathrm{H}=2.014102\mathrm{u};{}_3^6\mathrm{Li}=6.01513\mathrm{u};{}_3^7\mathrm{Li}=7.016004\mathrm{u}$; ${}_{64}^{152}\mathrm{Gd}=151.919803\mathrm{u};{}_{82}^{206}\mathrm{Pb}=205.974455\mathrm{u};{}_1^3\mathrm{H}=3.016050\mathrm{u};{}_2^4\mathrm{He}=4.002603\mathrm{u}$; ${}_{30}^{70}\mathrm{Zn}=69.925325\mathrm{u};{}_{34}^{82}\mathrm{Se}=81.919709\mathrm{u};{}_{81}^{205}\mathrm{Be}=204.974454\mathrm{u}]$

Show Answer

Correct answer: (3)

Solution:

Two light nuclei, of mass $m_1$ and $m_2$, can undergo fusion, to produce a nucleus of mass $M$, if

$${\mathrm{m}}_1+{\mathrm{m}}_2>\mathrm{M}$$

A heavy nucleus, of mass $M_1$, can break up into two nuclei of mass $m_3$ and $m_4$, if

${\mathrm{m}}_3+{\mathrm{m}}_4<{\mathrm{M}}_1$

(1) ${}_3^7\mathrm{Li}\to {}_2^4\mathrm{He}+{}_1^3\mathrm{H}$

Mass of ${}_3^7\mathrm{Li}={\mathrm{M}}_1=7.016004\mathrm{u}$

Sum of masses of ${}_2^4\mathrm{He}+{}_1^3\mathrm{H}={\mathrm{m}}_3+{\mathrm{m}}_4$

$$=4.002603+3.016050=7.018653\mathrm{u}$$

Since $(m_3+m_4)>M_1$; the reaction is not possible

(2) ${}_{82}^{206}\mathrm{Pb}\to {}_{81}^{205}\mathrm{Be}+{}_1^1\mathrm{H}$

$\text{ Mass of }{}_{82}^{206}\mathrm{Pb}={\mathrm{M}}_1=205.974454\mathrm{u}$

Sum of masses of ${}_{81}^{205}\mathrm{Be}+{}_1^1\mathrm{H}={\mathrm{m}}_3+{\mathrm{m}}_4$

$$=204.974454+1.007825=205.982279\mathrm{u}$$

Since $m_3+m_4>m_1$; reaction is not possible

(3) ${}_1^2\mathrm{H}+{}_2^4\mathrm{He}\to {}_3^6\mathrm{Li}$

Sum of mass of ${}_1^2\mathrm{H}+{}_2^4\mathrm{He}={\mathrm{m}}_1+{\mathrm{m}}_2$

$$=2.014102+4.002603=6.016705\mathrm{u}$$

Mass of ${}_3^6\mathrm{Li}={\mathrm{M}}_1=6.01513\mathrm{u}$

Since $m_1+m_2>M$, this reaction is possible

(4) ${}_{30}^{70}\mathrm{Zn}+{}_{34}^{82}\mathrm{Se}\to {}_{64}^{152}\mathrm{Gd}$

Sum of mass of ${}_{30}^{70}\mathrm{Zn}+{}_{34}^{82}\mathrm{Se}={\mathrm{m}}_1+{\mathrm{m}}_2$

$$=69.925325+81.916709=151.842034\mathrm{u}$$

Mass of ${}_{64}^{152}\mathrm{Gd}=\mathrm{M}=151.919803\mathrm{u}$

Since $(m_1+m_2)<M$, this reaction is not possible

Hence option (3) is correct.

Average

Nucleus

28. Figure shows a hypotetical binding energy per nucleon (B.E/A); versus (mass number) A curve. Which of the following statements is correct?

(1) Fusion of nuclei, having mass number in the range $20<\mathrm{A}<50$, will release energy

(2) Fission of nuclei, having mass number in the range $50<\mathrm{A}<100$, will release energy

(3) Fusion of nuclei, having mass number in the range $60<\mathrm{A}<100$, will release energy

(4) Fission of nuclei, having mass number in the range $125<\mathrm{A}<175$, will release energy, if the two fragments are identical

Show Answer

Correct answer: (3)

Solution:

Energy is released in a nuclear (fission or fusion) reaction, if the binding energy per nucleon of the products, is more than that for reactants. This is so only far statement (3). In all the other statements, B.E/nucleon, of products, is less than that of the reactants. In such a process energy, has to be absorbed.

Hence option(3) is correct.

Average

Nuclear Energy

29. In a process known as “pair-production” a $\gamma$-ray photon gets converted into an electronposition pair. The minimum frequency of the photon, for pair production to take place, is: $(\mathrm{m}=$ mass of electron $=$ mass of position $=9.1\times {10}^{-31}\mathrm{kg}$, Planck’s constant $=6.6\times {10}^{-34}\mathrm{J}-\mathrm{s})$

(1) $\sim 2.5\times {10}^{18}\mathrm{Hz}$

(2) $\sim 2\times {10}^{19}\mathrm{Hz}$

(3) $\sim 2.5\times {10}^{20}\mathrm{Hz}$

(4) $4\times {10}^{20}\mathrm{Hz}$

Show Answer

Correct answer: (3)

Solution:

The process of “pair production” is $\mathrm{h}v(\gamma -$ ray $)\to {}_{-1}^0\mathrm{e}+{}_{+1}^0\mathrm{e}$

The total mass energy of electron position pair

$$=2\left({\mathrm{mc}}^2\right)=2\times 9.1\times {10}^{-31}\times {(3\times {10}^8)}^2\mathrm{J}=163.8\times {10}^{-15}\mathrm{J}$$

For “pair production”, this must be the minimum energy of the $\gamma$-ray photon. Let $v_{\mathrm{m}}$ be the minimum frequency of such a $\gamma$-ray photon. Then

$$\begin{array}{rl}& \mathrm{h}{\nu }_{\mathrm{m}}=163.8\times {10}^{-15}\mathrm{J}\\ \therefore & {\nu }_{\mathrm{m}}={\displaystyle \frac{163.8\times {10}^{-15}}{6.6\times {10}^{-34}}}\mathrm{Hz}\simeq 2.482\times {10}^{20}\mathrm{Hz}\simeq 2.5\times {10}^{20}\mathrm{Hz}\end{array}$$

Hence option(3) is correct.

Difficult

Nuclear Energy

30. In a nuclear power generating station, ${}_{92}^{235}\mathrm{U}$ is used as “the fuel”. Fission of one atom of ${}_{92}^{235}\mathrm{U}$ releases $200\mathrm{MeV}$ of energy. The efficiency of generating plant is $80\mathrm{\%}$. The (reactor) plant consumes $256\mathrm{mg}$ of uranium fuel in 4 hours. Its power output, in MW is (nearly),

(1) 142

(2) 178

(3) 221

(4) 276

Show Answer

Correct answer: (2)

Solution:

Energy produced due to fission of one atom of ${}_{92}^{235}\mathrm{U}=200\mathrm{MeV}$

The number of atoms, in $256\mathrm{mg}$ of ${}_{92}^{235}\mathrm{U}={\displaystyle \frac{6.023\times {10}^{23}}{0.256\times 235}}\simeq {10}^{22}$

$\therefore$ Total energy produced due to fission of the given mass of ${}_{92}^{235}\mathrm{U}=200\times {10}^{22}\mathrm{MeV}=2\times {10}^{24}\mathrm{MeV}$

$$=2\times 1.6\times {10}^{-13}\times {10}^{24}\mathrm{J}=3.2\times {10}^{11}\mathrm{J}$$

The efficiency, $\eta ={\displaystyle \frac{E_{\text{out }}}{E_{\text{in }}}}=0.8$

$\therefore {\mathrm{E}}_{\text{out }}=0.8\times 3.2\times {10}^{11}\mathrm{J}\simeq 2.56\times {10}^{11}\mathrm{J}$

Time taken $=4\mathrm{hr}=4\times 3600\mathrm{s}$

$\therefore$ Power output $={\displaystyle \frac{2.56\times {10}^{11}}{4\times 36\times {10}^2}}\mathrm{J}/\mathrm{s}\simeq 1.78\times {10}^9\mathrm{W}=178\mathrm{MW}$

Hence option (2) is correct.

Average

Rutherford Scattering

31. In a Rutherfod scattering experiment when a projectile of charge $z_1$ and mass $m_1$ approaches a target nucleus of charge $z_2$ and mass $m_2$, the distance of closet approach is $r_0$. The energy of the projectile is

(1) directly proportional to $m_1$

(2) directly proportional to ${\mathrm{m}}_1{\mathrm{m}}_2$

(3) directly proportional to ${\mathrm{z}}_1{\mathrm{z}}_2$

(4) inversely proportional to ${\mathrm{z}}_1{\mathrm{z}}_2$

Correct answer: (3)

Solution:

When an particle is very far off it has only K.E (K). As it approaches the target its K.E decreases and electrostate P.E increases. At distance of closet approach $\left(r_0\right)$ the K.E. is zero. The entire energy is now electrostatic P.E. From law of conservation of energy

$$\mathrm{K}.\mathrm{E}={\displaystyle \frac{{\mathrm{z}}_1{\mathrm{Z}}_2}{4\pi {ϵ}_0{\mathrm{r}}_0}}$$

Hence option(3) is correct.

Average

Rutherford Scattering

32. An alpha nucleus of energy $\left(1/2{\mathrm{mv}}^2\right)$ bombards a heavy nuclear target of charge $\mathrm{Ze}$. Then the distance of closeset appraoch for the alpha to the nucleus is proportional to

(1) (1 / z e)

(2) $v^2$

(3) ${\displaystyle \frac{1}{\mathrm{m}}}$

(4) $\mathrm{m}$

Show Answer

Correct answer: (3)

Solution:

We know that the kinetic energy of alpha nucleus is equal to potential energy of alpha nucleus and target nucleus at the distance of closeset approach

$$\begin{array}{rl}& {\displaystyle \frac{1}{2}}{\mathrm{mv}}^2={\displaystyle \frac{{\mathrm{q}}_{\alpha }(\mathrm{Ze})}{4\pi {\epsilon }_0{\mathrm{r}}_0}}\\ & {\mathrm{r}}_0={\displaystyle \frac{2{\mathrm{q}}_{\alpha }(\mathrm{Ze})}{4\pi {\epsilon }_0{\mathrm{mv}}^2}}\end{array}$$

Hence ${\mathrm{r}}_0\propto {\displaystyle \frac{1}{\mathrm{m}}}$

Hence option (3) is correct.

Average

33. The binding energy of a hydrogen like atom in ground state is $217.6\mathrm{eV}$. The atomic number of the atom is

(1) 4

(2) 8

(3) 16

(4) 32

Show Answer

Correct answer: (1)

Solution:

The energy in ground state of a hydrogen like atom (atomic number $\mathrm{z}$ ) is proportional to ${\mathrm{z}}^2$. For the hydrogen atom this energy is $13.6\mathrm{eV}$ and $\mathrm{z}=1$

Hence ${\displaystyle \frac{217.6}{13.6}}={\displaystyle \frac{{\mathrm{z}}^2}{1}}$

or ${\mathrm{z}}^2=16$

Hence $\mathrm{z}=4$

Hence option (1) is correct

Difficult

Bohr’s Theory

34. In an atom two electrons move around the nucleus in circular orbits of radii $r$ and $4r$. The ratio of the time taken by then to complete one revolution around the nucleus is

(1) ${\displaystyle \frac{1}{16}}$

(2) ${\displaystyle \frac{1}{8}}$

(3) ${\displaystyle \frac{1}{2}}$

(4) 8

Show Answer

Correct answer: (2)

Solution:

If the electron is moving with a velocity $v$ around the nucleus the time taken by it to complete one revolution would be given by $\mathrm{T}={\displaystyle \frac{2\pi \mathrm{r}}{\mathrm{v}}}$

Now from Bohr’s theory we have

$$\mathrm{r}\propto {\mathrm{n}}^2\text{ and }\mathrm{v}\propto {\displaystyle \frac{1}{\mathrm{n}}}$$

Hence $\mathrm{T}\propto {\mathrm{n}}^3{\mathrm{T}}^2\propto {\mathrm{n}}^6$ or ${\mathrm{T}}^2\propto {\mathrm{r}}^3$

Hence we write

$${\left({\displaystyle \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}}\right)}^2={\left({\displaystyle \frac{\mathrm{R}}{4\mathrm{R}}}\right)}^3={\displaystyle \frac{1}{64}}\text{ or }{\displaystyle \frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}}={\displaystyle \frac{1}{8}}$$

Hence option (2) is correct

Average

Bohr’s Theory

35. According to Bohr’s theory, the minimum energy required to remove an electron from the ground state of doubly ionized ${\mathrm{L}}_{\mathrm{i}}$ atom $(\mathrm{Z}=3)$ is

(1) 1.51

(2) 13.6

(3) 40.8

(4) 122.4

Show Answer

Correct answer: (4)

Solution:

According to Bohr’s theory for hydrogen and hydrogen like atoms, the energy in electron volts is given by the relation.

$$\mathrm{E}=-13.6{\displaystyle \frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}}$$

For doublly ionised ${\mathrm{L}}_{\mathrm{i}}$ atom $\mathrm{z}=3$ and it will behave like a hydrogen atom. For ground state. $\mathrm{n}=1$ and hence we have

$$\mathrm{E}={\displaystyle \frac{-13.6\times (3{)}^2}{(1{)}^2}}\mathrm{eV}=122.4\mathrm{eV}$$

Hence option (4) is correct.

Difficult

Bohr’s Theory

36. The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\mathrm{A}$. The wave length of the second spectral line in the Balmer series of singly ionized helium atom is

(1) $1215\text{AA}$

(2) $1640\text{AA}$

(3) $2430\text{AA}$

(4) $4687\text{AA}$

Show Answer

Correct answer: (1)

Solution:

We know that Balmer series terminates at $\mathrm{n}=2$. Hence the first spectral line of Balmer series corresponds to transition from $n_1=3$ to $n_2=2$. Hence we have

$$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_{{\mathrm{H}}_1}}}=\mathrm{R}\times 1^2({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}})=\mathrm{R}({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{9}})\\ & {\lambda }_{{\mathrm{H}}_1}={\displaystyle \frac{36}{5\mathrm{R}}}\end{array}$$

or $6561={\displaystyle \frac{36}{5\mathrm{R}}}$

The second line of Balmer series of ionzied helium atom corresponds to transition from $n_1=4$ to $n_2=2$

Hence ${\displaystyle \frac{1}{{\lambda }_{\text{He }}(2)}}=\mathrm{R}\times 2^2({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{4^2}})$

$$\begin{array}{rl}& =4\mathrm{R}({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{6}})=4\mathrm{R}\left({\displaystyle \frac{3}{16}}\right)={\displaystyle \frac{3\mathrm{R}}{4}}\\ {\lambda }_{{\mathrm{He}}_{(2)}}& ={\displaystyle \frac{4}{3\mathrm{R}}}\\ {\displaystyle \frac{{\lambda }_{{\mathrm{He}}_{(2)}}}{{\lambda }_{{\mathrm{H}}_1}}}& ={\displaystyle \frac{4}{3\mathrm{R}}}\times {\displaystyle \frac{5\mathrm{R}}{36}}={\displaystyle \frac{5}{27}}\\ {\lambda }_{{\mathrm{He}}_{(2)}}& ={\displaystyle \frac{6561\times 5}{27}}=1215\text{AA}\end{array}$$

Hence option(1) is correct.

Difficult

Bohr’s Theory

37. Imagine an atom made up of a proton and a hypotheticle particle of double the mass of electron but having the same charge as that of the electron. Apply the Bohr atom model and consider all possible transition of the hypothetical particle to the first excited level. $R$ is the Rydberg constant. The longest wavelength photon that will be emitted has wavelength $\lambda$. In terms of $R$ the value of $\lambda$ will be

(1) $1.8\mathrm{R}$

(2) $7.2\mathrm{R}$

(3) $3.6\mathrm{R}$

(4) $4/\mathrm{R}$

Show Answer

Correct answer: (3)

Solution:

The energy expression corresponding to ${\mathrm{n}}^{\text{th }}$ permitted level is given by ${\mathrm{E}}_{\mathrm{n}}=-{\displaystyle \frac{\mathrm{Rhc}}{{\mathrm{n}}^2}}$

where $\mathrm{R}=$ Rydberg’s constant is directly proportional to mass $\mathrm{m}$ of electron.

Hence $m$ will get replaced by $2\mathrm{m}$

Also the energy expression can be written as $E_n=-{\displaystyle \frac{Rhc}{n^2}}$

Energy of the hypothetical particle will be ${\mathrm{E}}_{\mathrm{n}}=-{\displaystyle \frac{2\mathrm{Rhc}}{{\mathrm{n}}^2}}$

The longest wavelength ${\lambda }_{max}$ will be corresponding to minimum energy. This will happen when the hypothetical particle makes a transition from $n=3$ to $n=2$.

$$\therefore {\displaystyle \frac{\mathrm{hc}}{{\lambda }_{max}}}=2\mathrm{Rhc}({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}})$$

which gives ${\lambda }_{max}={\displaystyle \frac{18}{5\mathrm{R}}}=3.6\mathrm{R}$

Hence option(3) is correct.

Average

Bohr’s Theory

38. A muonic hydrogen atom is a bound state of a negatively charged muon of mass $207\mathrm{mg}$ and a proton and the meuon orbit around the proton. Then the ground state energy of mounic hydrogen atom is

(1) $-0.65\mathrm{eV}$

(2) $-2.8\times {10}^3\mathrm{eV}$

(3) $-13.6\mathrm{eV}$

(4) $-2.8\times {10}^2\mathrm{eV}$

Show Answer

Correct answer: (2)

Solution:

We known that the ground state energy of the hydrogen atom is gien by $E_1=-{\displaystyle \frac{2{\pi }^2{\mathrm{K}}^2{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^4}{{\mathrm{h}}^2}}$

Since the mass of muonic hydrogen atom is 207 times the mass of the electron, the ground state energy of this mounic atom would be given by

$$\begin{array}{rl}& \mathrm{E}(\text{ mounic })={\mathrm{E}}_1=-{\displaystyle \frac{2{\pi }^2{\mathrm{K}}^2{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^4}{{\mathrm{h}}^2}}\times 207=-13.6\times 207\mathrm{eV}\\ & =-2.8\times {10}^3\mathrm{eV}\end{array}$$

Hence option (2) is correct.

Average

Bohr’s Thoery

39. The first excitation potential of hydrogen is

(1) $13.6\mathrm{V}$

(2) $10.22\mathrm{V}$

(3) $12.1\mathrm{V}$

(4) $1.89\mathrm{V}$

Show Answer

Correct answer: (2)

Solution:

The first excitation potential of hydrogen atom is the potential for transition from Ist Bohr’s orbit to the second Bohr orbit.

Hence $V={\displaystyle \frac{E_2-E_1}{e}}={\displaystyle \frac{hc}{e}}R({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}})$

$$\mathrm{V}={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}}=10.22\mathrm{V}$$

Hence option(2) is correct.

Difficult

Bohr’s Thoery

40. The transition from the state $n=4$ to $n=3$ in a hydrogen atom results in ultraviolet radiation. Infrared radiation will be obtained in transition.

(1) $2\to 1$

(2) $3\to 2$

(3) $4\to 2$

(4) $5\to 4$

Show Answer

Correct answer: (4)

Solution:

Energy of infrared radiation is less than the energy of ultraviolet radiation. As shown below the energy released will be less in option (4) while the energy released will be more in option (1), (2) and (3). Hence option (4) is correct.

For transition $4\to 3\to {\mathrm{E}}_2-{\mathrm{E}}_1=\mathrm{R}({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{4^2}})$ ultraviolet

$$=\mathrm{R}({\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{16}})={\displaystyle \frac{7}{128}}\mathrm{R}=(0.0546)\mathrm{R}$$

For transition $2\to 1\to ({\mathrm{E}}_2-{\mathrm{E}}_1)=\mathrm{R}(1-{\displaystyle \frac{1}{4}})={\displaystyle \frac{3}{4}}\mathrm{R}=0.75\mathrm{R}$

For transition $3\to 2\to ({\mathrm{E}}_2-{\mathrm{E}}_1)=\mathrm{R}({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{9}})={\displaystyle \frac{5}{36}}\mathrm{R}=0.088\mathrm{R}$

For transition $4\to 2\to \mathrm{R}({\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{16}})={\displaystyle \frac{3}{16}}\mathrm{R}=0.1875\mathrm{R}$

For transition $5-4\to R({\displaystyle \frac{1}{16}}-{\displaystyle \frac{1}{25}})={\displaystyle \frac{9}{400}}R=0.225R$

From this data we find that option (4) is the correct option.

Average

Bohr’s Theory

41. An excited hydrogen atom emits a photon of wavelength $\lambda$ when it jumps to ground state from the excited state say of quantum number $n$. If $R$ is the Rydberg constant the value of $\mathbf{n}$ is given by

(1) $\sqrt{{\displaystyle \frac{\lambda \mathrm{R}}{\lambda (\mathrm{R}-1)}}}$

(2) $\sqrt{{\displaystyle \frac{\lambda }{(\lambda \mathrm{R}-1)}}}$

(3) $\sqrt{{\displaystyle \frac{(\lambda \mathrm{R}-1)}{\lambda \mathrm{R}}}}$

(4) $\sqrt{{\displaystyle \frac{(\lambda \mathrm{R}-1)}{\mathrm{R}}}}$

Show Answer

Correct answer: (1)

Solution:

According to Bohr’s theory we have $\overline{v}={\displaystyle \frac{1}{\lambda }}=R({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{n^2}})$

or ${\displaystyle \frac{\mathrm{R}}{{\mathrm{n}}^2}}=(\mathrm{R}-{\displaystyle \frac{1}{\lambda }})={\displaystyle \frac{\lambda \mathrm{R}-1}{\lambda }}$

$\begin{array}{rl}& {\mathrm{n}}^2={\displaystyle \frac{\lambda \mathrm{R}}{(\lambda \mathrm{R}-1)}}\\ & \mathrm{n}=\sqrt{{\displaystyle \frac{\lambda \mathrm{R}}{(\lambda \mathrm{R}-1)}}}\end{array}$

Hence option(1) is correct.

Difficult

Bohr’s Theory

42. Hydrogen atom is excited from ground state to a state corresponding to principal quantum number 4. The number of spectral lines obtained in the emission spectra is given by

(1) 2

(2) 3

(3) 5

(4) 6

Show Answer

Correct answer: (4)

Solution:

The number of lines emitted in an emission spectra of Hydrogen corresponding to a transition from a state of quantum number $n$ to ground state is given by ${\displaystyle \frac{n(n-1)}{2}}$

$\therefore$ The number of lines that would be emitted when a transition take place from $n=4$ to ground state is given by ${\displaystyle \frac{4(4-1)}{2}}=6$

Hence option(4) is correct.

Average

Bohr’s Theory

43. In a hydrogen like atom, electron makes transition from an energy level with quantum number $n$ to another quantum number ( $n-1$ ). If $n»1$, the frequency of the radiation emitted is proportional to

(1) ${\displaystyle \frac{1}{\mathrm{n}}}$

(2) ${\displaystyle \frac{1}{{\mathrm{n}}^2}}$

(3) ${\displaystyle \frac{1}{{\mathrm{n}}^{3/2}}}$

(4) ${\displaystyle \frac{1}{{\mathrm{n}}^3}}$

Show Answer

Correct answer: (4)

Solution:

According to Bohr’s theory of hydrogen atom, frequency of radiation emitted is given by relation

$$\mathrm{h}\nu =\mathrm{R}[{\displaystyle \frac{1}{(\mathrm{n}-1{)}^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}}]$$

where $\mathrm{R}$ is Rydberg like constant for hydrogen like atom

$$\begin{array}{rl}\nu & ={\displaystyle \frac{1}{h}}R\left[{\displaystyle \frac{n^2-(n-1{)}^2}{n^2(n-1{)}^2}}\right]\\ \text{ or }\nu & ={\displaystyle \frac{R}{h}}\left[{\displaystyle \frac{(n+n-1)(n+n+1)}{n^2(n-1{)}^2}}\right]\\ \nu & ={\displaystyle \frac{R}{h}}\left[{\displaystyle \frac{2n-1}{n^2(n-1{)}^2}}\right]\end{array}$$

Since $\mathrm{n}»1$ we have $\nu \simeq {\displaystyle \frac{\mathrm{R}}{\mathrm{h}}}\left({\displaystyle \frac{2\mathrm{n}}{{\mathrm{n}}^4}}\right)={\displaystyle \frac{\mathrm{R}}{\mathrm{h}}}\left({\displaystyle \frac{1}{{\mathrm{n}}^3}}\right)$

or $\nu \propto {\displaystyle \frac{1}{{\mathrm{n}}^3}}$

Hence option(4) is correct.

Difficult

Bohr’s Theory

44. Electrons in hydrogen atom first jump from third excited state to second excited state and then from second excited state to the first excited state. The ratio of ${\lambda }_1$ and ${\lambda }_2$ emitted in the two cases is

(1) ${\displaystyle \frac{7}{5}}$

(2) ${\displaystyle \frac{27}{20}}$

(3) ${\displaystyle \frac{27}{5}}$

(4) ${\displaystyle \frac{20}{7}}$

Show Answer

Correct answer: (4)

Solution:

For transition ${\mathrm{n}}_2=3$ to ${\mathrm{n}}_1=2$ we have

$${\displaystyle \frac{\mathrm{hc}}{{\lambda }_1}}={\mathrm{E}}_3-{\mathrm{E}}_4=13.6[{\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{4^2}}]=13.6\times {\displaystyle \frac{1}{44}}$$

For transitionto ${\mathrm{n}}_2=2$ to ${\mathrm{n}}_1=1$ we similarly have

$${\displaystyle \frac{\mathrm{hc}}{{\lambda }_2}}={\mathrm{E}}_2-{\mathrm{E}}_3=13.6[{\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}}]$$

From above two equations we have

$${\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}={\displaystyle \frac{5/36}{7/144}}={\displaystyle \frac{20}{7}}$$

Hence option (4) is correct.

Difficult

Bohr’s Theory

45. The binding energy of a hydrogen like atom in ground state is $217.6\mathrm{eV}$. An electron having energy of $130\mathrm{eV}$ is incident on the atom. Then the energy of the electron emerging after interacting with this atom is

(1) $217.6\mathrm{eV}$

(2) $87.6\mathrm{eV}$

(3) $130\mathrm{eV}$

(4) $347.6\mathrm{eV}$

Show Answer

Correct answer: (3)

Solution:

The electron undergoes an elastic collision with the concerned atom. Since $130\mathrm{eV}$ cannot raise the electron from ground state to the first excited stae, the electron does not accept any energy from the electron. Hence the electron will emerge after interaction with the atom with $130\mathrm{eV}$.

Hence option(3) is correct.

Difficult

Bohr’s Theory

46. The different lines in Lyman’s series have their wavelength lying between

(1) 0 to $\mathrm{\infty }$

(2) $911\text{AA}$ to $1215\text{AA}$

(3) $1000\text{AA}$ to $1500\text{AA}$

(4) $915\text{AA}$ to $1211\text{AA}$

Show Answer

Correct answer: (2)

Solution:

For Lyman series, the wavelength of lines is obtained from the relation $\overline{v}={\displaystyle \frac{1}{\lambda }}=\mathrm{R}({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}})$

We will get a line with shortest wavelength corresponding to transition from $n=\mathrm{\infty }$ to $n=1$

Hence ${\displaystyle \frac{1}{{\lambda }_{\mathrm{s}}}}=\mathrm{R}({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{{\mathrm{n}}^2}})=\mathrm{R}$

$${\lambda }_{\mathrm{S}}={\displaystyle \frac{1}{\mathrm{R}}}={\displaystyle \frac{{10}^{10}}{10967700}}=911\text{AA}$$

For longest wavelength we will have $\mathrm{n}=2$

Hence ${\lambda }_{\mathrm{L}}=\mathrm{R}(1-{\displaystyle \frac{1}{4}})={\displaystyle \frac{3}{4}}\mathrm{R}$

$$\lambda ={\displaystyle \frac{4}{3}}\mathrm{R}={\displaystyle \frac{4\times {10}^{10}}{3\times 10967700}}=1215\text{AA}$$

Hence option (2) is correct.

Difficult

Bohr’s Theory

47. The ratio of wavelength ${\mathbf{H}}_{\alpha }$ and ${\mathbf{H}}_{\beta }$ lines of Paschen’s series is of the order of

(1) 10

(2) 0.1

(3) 1.5

(4) 100

Show Answer

Correct answer: (3)

Solution:

The Pachen’s series is generated when transition from higher energy levels terminates at ${\mathrm{n}}_1=3$. Now ${\mathrm{H}}_{\alpha }$ line of Paschen’s series will correspond to a transition for $4^{\text{th }}$ to $3^{\text{rd }}$ energy level. Hence

$${\overline{v}}_{\alpha }={\displaystyle \frac{1}{{\lambda }_{\alpha }}}=\mathrm{R}({\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{4^2}})=\mathrm{R}({\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{16}})$$

or ${\lambda }_{\alpha }={\displaystyle \frac{144}{7\mathrm{R}}}$

Similarly ${\mathrm{H}}_{\beta }$ line will be obtained corresponding to ${\mathrm{n}}_1=3$ and ${\mathrm{n}}_2=5$

$$\begin{array}{rl}& {\overline{v}}_{\beta }={\displaystyle \frac{1}{{\lambda }_{\beta }}}=\mathrm{R}({\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{5^2}})=\mathrm{R}({\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{25}})={\displaystyle \frac{16}{225}}\mathrm{R}\\ & {\lambda }_{\beta }={\displaystyle \frac{225}{16\mathrm{R}}}\\ & {\displaystyle \frac{{\lambda }_{\alpha }}{{\lambda }_{\beta }}}={\displaystyle \frac{144}{7\mathrm{R}}}\times {\displaystyle \frac{16\mathrm{R}}{225}}\approx 1.5\end{array}$$

Hence option (3) is correct.

Difficult

Bohr’s Theory

48. The ratio of minimum to maximum wavelength in Balmer series is

(1) $5:36$

(2) $36:5$

(3) $9:5$

(4) $5:9$

Show Answer

Correct answer: (4)

Solution:

Since the series involved is Balmer series, the transition terminates at $\mathrm{n}=2$

$$\therefore \overline{v}={\displaystyle \frac{1}{\lambda }}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$$

It is clear that $\lambda$ will have maximum value corresponding to transition from $n_2=3$ to $n_1=2$

$$\therefore {\displaystyle \frac{1}{{\lambda }_{max}}}=\mathrm{R}({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}})=\mathrm{R}\left({\displaystyle \frac{5}{36}}\right)$$

For minimum value of $\lambda$ the transition has to be from $\mathrm{\infty }$ to $n_1=2$

$$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_{min}}}=\mathrm{R}({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{\mathrm{\infty }}})=\mathrm{R}{\displaystyle \frac{1}{4}}\\ & {\displaystyle \frac{{\lambda }_{min}}{{\lambda }_{max}}}={\displaystyle \frac{4}{\mathrm{R}}}\times {\displaystyle \frac{5\mathrm{R}}{36}}={\displaystyle \frac{5}{9}}\end{array}$$

Hence option(4) is correct.

Difficult

Bohr’s Theory

49. The ratio of the longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is

(1) ${\displaystyle \frac{9}{31}}$

(2) ${\displaystyle \frac{5}{27}}$

(3) ${\displaystyle \frac{3}{23}}$

(4) ${\displaystyle \frac{7}{29}}$

Show Answer

Correct answer: (2)

Solution:

According to the relation $\overline{v}={\displaystyle \frac{1}{\lambda }}=\mathrm{R}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$ the longest wavelength is obtained when the transition take place froma quantum state one higher than the quantum state where the series terminates. Hence for Lyman series

$${\displaystyle \frac{1}{{\lambda }_{\mathrm{L}}}}=\mathrm{R}({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{2^2}})=\mathrm{R}\left({\displaystyle \frac{3}{4}}\right)$$

and for Balmer series

$$\begin{array}{rl}& {\displaystyle \frac{1}{{\lambda }_{\mathrm{B}}}}=\mathrm{R}({\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{3^2}})=\mathrm{R}\left({\displaystyle \frac{5}{36}}\right)\\ & {\displaystyle \frac{{\lambda }_{\mathrm{L}}}{{\lambda }_{\mathrm{B}}}}={\displaystyle \frac{5\mathrm{R}}{36}}\times {\displaystyle \frac{4}{3\mathrm{R}}}={\displaystyle \frac{5}{27}}\end{array}$$

Hence option(2) is correct.

Difficult

Bohr’s Theory

50. An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the electron acquired as a result of photon emission will be

(1) ${\displaystyle \frac{24\mathrm{m}}{25\mathrm{hR}}}$

(2) ${\displaystyle \frac{24\mathrm{hR}}{25\mathrm{m}}}$

(3) ${\displaystyle \frac{25\mathrm{m}}{24\mathrm{hR}}}$

(4) ${\displaystyle \frac{24\mathrm{m}}{25\mathrm{hR}}}$

Show Answer

Correct answer: (2)

Solution:

The wavelength of the emitted radiation when the transition take place from $n_2=5$ to $n_1=1$ is given by the relation

$$\overline{v}={\displaystyle \frac{1}{\lambda }}\mathrm{R}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$$

$${\displaystyle \frac{1}{\lambda }}=\mathrm{R}({\displaystyle \frac{1}{1^2}}-{\displaystyle \frac{1}{5^2}})={\displaystyle \frac{24\mathrm{R}}{25}}\text{ or }\lambda ={\displaystyle \frac{25}{24\mathrm{R}}}$$

Linear momentum of photon emitted is $p={\displaystyle \frac{h}{\lambda }}={\displaystyle \frac{h\times 24R}{25}}$

Now $\mathrm{p}=\mathrm{m}\nu (\because$ hydrogen atom is stationary $)$

$$\mathrm{m}\nu ={\displaystyle \frac{24\mathrm{Rh}}{25}}\nu ={\displaystyle \frac{24\mathrm{hR}}{25\mathrm{M}}}$$

Hence option (2) is correct.

Difficult

Bohr’s Theory

51. A diatomic molecule is made up of two masses $m_1$ and $m_2$ which are seperated by a distance $r$. If we calculate the rotational energy by applying Bohr’ hypothesis of angular momentum quantization its energy will be given by

(1) ${\displaystyle \frac{({\mathrm{m}}_1+{\mathrm{m}}_2){\mathrm{n}}^2{\mathrm{h}}^2}{2{\mathrm{m}}_1^2{\mathrm{m}}_2^2{\mathrm{r}}^2}}$

(2) ${\displaystyle \frac{n^2{\mathrm{h̸}}^2}{2(m_1+m_2)r^2}}$

(3) ${\displaystyle \frac{2n^2{h}^2}{(m_1+m_2)r^2}}$

(4) ${\displaystyle \frac{({\mathrm{m}}_1+{\mathrm{m}}_2){\mathrm{n}}^2{\mathrm{h̸}}^2}{2{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{r}}^2}}$

Show Answer

Correct answer: (4)

Solution:

Bohr’s condition for quantization of angular momentum is $\mathrm{mvr}={\displaystyle \frac{\mathrm{nh}}{2\pi }}$

where $\mathrm{m}$ is the reduced mass of diatomic molecule and is $={\displaystyle \frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_1{\mathrm{m}}_2}}$

If $\mathrm{v}$ denotes linear velocity and $\omega$ the angular velocity we have $\mathrm{v}=\mathrm{r}\omega$

$\therefore \mathrm{m}(\mathrm{r}\omega )\mathrm{r}=\mathrm{n}\hslash$ or $\omega ={\displaystyle \frac{\mathrm{n}\hslash }{{\mathrm{mr}}^2}}$

The rotational energy of the given diatomic molecule is given by

$${\displaystyle \frac{1}{2}}\mathrm{I}{\omega }^2={\displaystyle \frac{1}{2}}\left({\mathrm{mr}}^2\right){\left({\displaystyle \frac{\mathrm{n}\hslash }{{\mathrm{mr}}^2}}\right)}^2$$

$\therefore$ Rotational energy is ${\displaystyle \frac{{\mathrm{n}}^2{\hslash }^2}{2{\mathrm{mr}}^2}}={\displaystyle \frac{{\mathrm{n}}^2{\hslash }^2({\mathrm{m}}_1+{\mathrm{m}}_2)}{2{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{r}}^2}}$

Hence option (4) is correct.

Difficult

Bohr’s Theory

52. The electron in the hydrogen atom jumps from excited state $n=3$ to ground state $n=1$. The photon thus emitted radiate a photosensitive material. If the work function of the material is $5.1\mathrm{eV}$ the stopping potential of the photosensitive material is given by

(1) $5.1\mathrm{V}$

(2) $17.2\mathrm{V}$

(3) $7\mathrm{V}$

(4) $12.1\mathrm{V}$

Show Answer

Correct answer: (3)

Solution:

When a transition take place from state $\mathrm{n}=3$ to $\mathrm{n}=1$, the energy of photon emitted is

$$\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1={\displaystyle \frac{-13.6}{3^2}}-\left({\displaystyle \frac{-13.6}{1^2}}\right)=12.1\mathrm{eV}\text{ which equals to h}v$$

Let $V_s$ be the stopping potential of the photo sensitive material. Then according to Einstein’s photoelectric equation we must have

$${\mathrm{eV}}_{\mathrm{s}}=\mathrm{h}\nu -{\omega }_0=(12.1-5.1)=7.0\mathrm{eV}$$

or ${\mathrm{V}}_{\mathrm{s}}=7\mathrm{V}$

Hence option (3) is correct.

Difficult

Bohr’s Theory

53. A doubly ionized lithium atom can be considered as a hydrogen like atom with atomic number 3. An electron is to be excited from the first to third Bohr’s orbit (ionizon energy of hydrogen atom is $13.6\mathrm{eV}$ ). The wavelength of the radiation required to achieve this excitation is

(1) $0.114\times {10}^{-6}\mathrm{m}$

(2) $114.26\mathrm{A}$

(3) $0.114\mathrm{A}$

(4) $1142\mathrm{A}$

Show Answer

Correct answer: (2)

Solution:

Th energy of the electron in $n^{\text{th }}$ Bohr’s orbit with atomic number $z$ is given by $z^2$ ( ${\mathrm{E}}_{\mathrm{n}}$ of hydrogen)

Energy ${\mathrm{E}}_1$ for lithium ${}^{++}$in ${\mathrm{I}}^{\mathrm{st}}$ Bohr orbit is $-(13.6)\times 9\mathrm{eV}$

Energy ${\mathrm{E}}_3$ for lithium ${}^{++}$in $3^{\text{rd }}$ Bohr orbit is $-{\displaystyle \frac{13.6\times 9}{(3{)}^2}}\mathrm{eV}=-13.6\mathrm{eV}$

$\therefore$ Energy required for excitation from $\mathrm{n}=1$ to $\mathrm{n}=3$ for lithium ${}^{++}$

$$E_3-E_1=13.6(9-1)=13.6\times 8\mathrm{eV}=13.6\times 8\times 1.6\times {10}^{-19}\mathrm{J}$$

The wavelength $\lambda$ of radiation requird for this excitation to happen is

$$\lambda ={\displaystyle \frac{\mathrm{hc}}{{\mathrm{E}}_3-{\mathrm{E}}_1}}={\displaystyle \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{13.6\times 8\times 1.6\times {10}^{-19}}}=0.11426\times {10}^{-7}\mathrm{m}=114.26\mathrm{A}$$

Hence option(2) is correct.

Difficult

Bohr’s Theory

54. The ionisation energy of the hydrogen atom is $13.6\mathrm{eV}$. A photon is incident on a hydrogen atom which is intially in the ground state and excite it to the $n=4$ state. Then the wavelength of the photon is (Given $h=4.14\times {10}^{-15}\mathrm{eV}$ )

(1) $974\text{AA}$

(2) $974\times {10}^{-6}\mathrm{m}$

(3) $974\times {10}^{-7}\mathrm{m}$

(4) $974\times {10}^{-3}\mathrm{m}$

Show Answer

Correct answer: (1)

Solution:

We know the energy of an electron in th $n^{\text{th }}$ orbit of hydrogen atom is given by $E_n=-{\displaystyle \frac{13.6}{n^2}}\mathrm{eV}$ for $\mathrm{n}=1$ (the ground state) ${\mathrm{E}}_1=-13.6\mathrm{eV}$

for $\mathrm{n}=4{\mathrm{E}}_4=-{\displaystyle \frac{13.6}{16}}=-0.85\mathrm{eV}$

Energy of photon $={\mathrm{E}}_4-{\mathrm{E}}_1=-0.85-(-13.6)=12.75\mathrm{eV}$ and this must be equal to ${\displaystyle \frac{\mathrm{hc}}{\lambda }}$

$$\begin{array}{rl}& {\displaystyle \frac{\mathrm{hc}}{\lambda }}=12.75\mathrm{eV}\\ & \lambda ={\displaystyle \frac{\mathrm{hc}}{12.75}}={\displaystyle \frac{4.14\times {10}^{-15}\times 3\times {10}^8}{12.75}}=974\mathrm{A}\end{array}$$

Hence option(1) is correct.

Difficult

Bohr’s Theory

55. A gas of identical hydrogen like atoms has some atoms in the lowest ground energy level $A$ and some in a particular upper (excited) energy level and there are no atoms in any other energy level. The atoms of the gas move by transition to a higher energy level by absorbing monochromatic light of photon energy $2.7\mathrm{eV}$. Subsequently the atoms emit radiation of only six different photon energy. Some of the emitted photons have energy $2.7\mathrm{eV}$. Some have energy more and some have less than $2.7\mathrm{eV}$. This data is used to find the principal quantum number of the initially excited level $B$. This qunatum number is

(1) 3

(2) 4

(3) 1

(4) 2

Show Answer

Correct answer: (1)

Solution:

Let $\mathrm{z}$ be the atomic number of hydrogen like atom. Refering to figure we find that

${\mathrm{E}}_2-{\mathrm{E}}_1={\mathrm{z}}^2(10.2\mathrm{eV})$

${\mathrm{E}}_3-{\mathrm{E}}_1={\mathrm{z}}^2(12.1\mathrm{eV})$

${\mathrm{E}}_4-{\mathrm{E}}_1={\mathrm{z}}^2(12.75\mathrm{eV})$

Also ${\mathrm{E}}_3-{\mathrm{E}}_2={\mathrm{z}}^2(1.89\mathrm{eV})$

${\mathrm{E}}_4-{\mathrm{E}}_3={\mathrm{z}}^2(0.66\mathrm{eV})$

If the $2.7\mathrm{eV}$ photon circle the atom, then state of quantum number 1 and 2 have to be excluded because

$$2.7<{\mathrm{z}}^2=({\mathrm{E}}_2-{\mathrm{E}}_1)<{\mathrm{z}}^2=({\mathrm{E}}_3-{\mathrm{E}}_2)$$

It is possible that $2.7\mathrm{eV}$ photon may excite the atom from $3^{\text{rd }}$ to $4^{\text{th }}$ orbit. Then $2.7\mathrm{eV}={\mathrm{z}}^2(0.66\mathrm{eV})$ which gives $\mathrm{z}=2$

Hence the quantum number of intially excited state $B$ is $n=3$

Hence option(1) is correct.

Average

Critical Potential

56. A critical potential for hydrogen is $13.05\mathrm{V}$. A hydrogen atom is bombarded by an electron of corresponding energy. Then the wavelength of radiation emitted by the hydrogen atom is

(1) $9515\text{AA}$

(2) $95.15\times {10}^{-6}\mathrm{m}$

(3) $9515\times {10}^{-12}\mathrm{m}$

(4) $9515\times {10}^{-4}\mathrm{m}$

Show Answer

Correct answer: (1)

Solution:

Let $\lambda$ be required wavlength

Then we have $\mathrm{eV}={\displaystyle \frac{\mathrm{hc}}{\lambda }}$ where $\mathrm{V}$ denotes the corresponding critical potential

$\therefore \lambda ={\displaystyle \frac{\mathrm{hc}}{\mathrm{eV}}}={\displaystyle \frac{6.61\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}\times 13.05}}=9515\mathrm{A}$

Hence option (4) is correct.

57. Beta rays emitted by a radioactive material are

(1) electromagnetic radiations

(2) the electrons orbiting around the nucleus

(3) charged particles emitted by the nucleus

(4) neutral particle

Show Answer

Correct answer: (3)

Solution:

From the basic studies of the phenomenon of radioactivity, it is known that $\beta$ particles are fast moving electrons emitted by the nucleus. One can confirm this by determing the $\mathrm{e}/\mathrm{m}$ values of these particles.

Hence option(3) is correct.

Average

Radioactivity

58. Which of the follwing processes represent a $\gamma$ decay

(1) ${}_Z^A\mathrm{X}+\gamma \to {}_Z^A\mathrm{X}+\mathrm{a}+\mathrm{b}$

(2) ${}_Z^A\mathrm{X}+{}_0^1\mathrm{n}\to {}_{\mathrm{Z}-2}^{\mathrm{A}-3}\mathrm{X}+\mathrm{c}$

(3) ${}_Z^{A}X\to {}_Z^{A}X+f$

(4) ${}_Z^{A}X+{}_{-1}^0\mathrm{e}\to {}_{Z-1}^{A}X+g$

Show Answer

Correct answer: (3)

Solution:

During $\gamma$ decay atomic number (z) and mass number A does not change. In option (3), this is the case. In other options, either A or z, or both, are changing.

Hence option (3) is correct.

Average

Radioactivity

59. Half life of a radioactive substance $A$ is 4 day.s The probability that a nucleus will decay in two half lives is

(1) ${\displaystyle \frac{1}{4}}$

(2) ${\displaystyle \frac{3}{4}}$

(3) ${\displaystyle \frac{1}{16}}$

(4) 1

Show Answer

Correct answer: (2)

Solution:

After two half lives $(1/4{)}^{\text{th }}$ fraction of nuclei will remain unchanged. Hence $(3/4{)}^{\text{th }}$ fraction will decay. Hence the probability that a nucleus decays in two half lives is ${\displaystyle \frac{3}{4}}$.

Hence option(2) is correct.

Difficult

Radioactivity

60. A radioactive nucleus can decay by two difficient modes. The half life for the first mode is $T_1$ and that for the second mode is $T_2$. The effective half life $T$ of the nucleus would be

(1) $\mathrm{T}={\mathrm{T}}_1+{\mathrm{T}}_2$

(2) ${\displaystyle \frac{1}{\mathrm{T}}}={\displaystyle \frac{1}{{\mathrm{T}}_1}}+{\displaystyle \frac{1}{{\mathrm{T}}_2}}$

(3) $\ell n\left({\displaystyle \frac{1}{\mathrm{T}}}\right)=\ln \left({\displaystyle \frac{1}{{\mathrm{T}}_1}}\right)+\ln \left({\displaystyle \frac{1}{{\mathrm{T}}_2}}\right)$

(4) ${\displaystyle \frac{1}{\mathrm{T}}}={\displaystyle \frac{1}{({\mathrm{T}}_1+{\mathrm{T}}_2)}}$

Show Answer

Correct answer: (2)

Solution:

Denoting by ${\lambda }_1$ and ${\lambda }_2$ the decay constants of the two modes by which nucleus can decay we have

$${\lambda }_1={\displaystyle \frac{0.693}{{\mathrm{T}}_1}}\text{ and }{\lambda }_2={\displaystyle \frac{0.693}{{\mathrm{T}}_2}}$$

The probability that an active nucleus decay by first mode in a small time dt is ${\lambda }_1\mathrm{dt}$ and that for the second mode is ${\lambda }_2\mathrm{dt}$. Then the probability that it decays either by first process or second process is $({\lambda }_1\mathrm{dt}+{\lambda }_2\mathrm{dt})$. Now we can say that $\lambda$ is the effective decay constant. Hence the probability that the nucleus decay by effective value of $\lambda$ in small time dt is $\lambda \mathrm{dt}$.

Hence we must have

$$\lambda \mathrm{dt}={\lambda }_1\mathrm{dt}+{\lambda }_2\mathrm{dt}\text{ or }\lambda ={\lambda }_1+{\lambda }_2$$

or ${\displaystyle \frac{0.693}{\mathrm{T}}}={\displaystyle \frac{0.693}{{\mathrm{T}}_1}}+{\displaystyle \frac{0.693}{{\mathrm{T}}_2}}$

or ${\displaystyle \frac{1}{\mathrm{T}}}={\displaystyle \frac{1}{{\mathrm{T}}_1}}+{\displaystyle \frac{1}{{\mathrm{T}}_2}}$

Hence option (2) is correct.

Difficult

Radioactivity

61. To determine the half life of radioactive element, a graph is plotted between $\ell n\left|{\displaystyle \frac{dN}{dt}}\right|$ versus $t\left|{\displaystyle \frac{dN}{dt}}\right|$ representing the rate of radioactive decay at time $t$. If the number of radioactive nuclei of this element decreases by a factor of $p$ after 4.15 years, the value of $p$ is

(1) 2

(2) 4

(3) 6

(4) 8

Show Answer

Correct answer: (4)

Solution:

$\because \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\lambda \mathrm{t}}$ hence $\left|{\displaystyle \frac{\mathrm{dN}}{\mathrm{dt}}}\right|=\left|\lambda {\mathrm{N}}_0{\mathrm{e}}^{-\lambda \mathrm{t}}\right|$

$$\begin{array}{rl}& \ell n\left({\displaystyle \frac{dN}{dt}}\right)=\ell n_e\left(\lambda N_0\right)-\lambda t\\ & \ell n\left|{\displaystyle \frac{dN}{dt}}\right|=-\lambda t+{\log }_e\left(\lambda N_0\right)\end{array}$$

which is of the form of a straight line $y=mx+c$ whose slope is $-\lambda ={\displaystyle \frac{3-4}{6-4}}$

or $\lambda =\left({\displaystyle \frac{1}{2}}\right)$ year ${}^{-1}$

Half life $\mathrm{T}={\displaystyle \frac{0.693}{\lambda }}=2\times 0.693=1.386$ years

Now $\mathrm{N}={\mathrm{N}}_0{\left({\displaystyle \frac{1}{2}}\right)}^{1/2}$ or ${\displaystyle \frac{\mathrm{N}}{{\mathrm{N}}_0}}={\left({\displaystyle \frac{1}{2}}\right)}^{1/T}$

$$\begin{array}{rl}{\displaystyle \frac{1}{p}}& =\left({\displaystyle \frac{1}{2}}\right){\displaystyle \frac{4.16}{1.386}}={\left({\displaystyle \frac{1}{2}}\right)}^3={\displaystyle \frac{1}{8}}\\ \text{ or }p& =8\end{array}$$

Hence option (4) is correct.

Easy

Radioactivity

62. 10 milligrams of a radioactive material of half life two years are kept in a store for four years. Then the amount of the material remaining unchanged is

(1) $\sim 7.5\mathrm{mg}$

(2) $\sim 2.5\mathrm{mg}$

(3) $\sim 0.25\mathrm{mg}$

(4) $\sim 0$

Show Answer

Correct answer: (2)

Solution:

Half life of the given material $={\tau }_{1/2}=2\mathrm{yr}$

The time interval $=\mathrm{t}=4\mathrm{yr}$. (given) $=2{\tau }^{1/2}$

The amount of mass remaining $={\displaystyle \frac{10\mathrm{mg}}{(2{)}^2}}=2.5\mathrm{mg}$

Hence option (2) is correct.

Average

Radioactivity

63. For a given sample, the counting rate is $47.5\alpha$ particle/ $\min$. After 5 minutes the count is reduced to $27\alpha$ particles $/\min$. Then the half life of the sample is

(1) 11 minute

(2) 5.99 minute

(3) 0.59 minute

(4) 0.11 minute

Show Answer

Correct answer: (2)

Solution:

Initial counting rate $=47.5\mathrm{a}$ particles $/\min$

Final counting rate after 5 minutes $27\alpha$ particles/min

$\therefore 27=47.5{\mathrm{e}}^{-\lambda }$ or ${\displaystyle \frac{47.5}{27}}={\mathrm{e}}^{\lambda }$

or $\lambda ={\displaystyle \frac{2.303\times 0.2455}{5}}=0.1157{\min }^{-1}$

Half life of sample ${\mathrm{T}}_{1/2}={\displaystyle \frac{0.693}{\lambda }}={\displaystyle \frac{0.693}{0.1157}}\min =5.989\min \simeq 5.99$ minute

Hence option (2) is correct.

Difficult

Radioactivity

64. One milligram of thorium emits $22\alpha$ particlea per unit solid angle per minute. The half life of thorium is of the order of

(1) $\sim 100$ years

(2) $\sim 1000$ years

(3) $\sim 10000$ years

(4) $\sim {10}^{10}$ years

Show Answer

Correct answer: (4)

Solution:

Number of $\alpha$ particle per sec in all direction is given by ${\displaystyle \frac{22\times 4\pi }{60}}={\displaystyle \frac{22\pi }{15}}$

No. of atoms in 1 milligram of thorium $={\displaystyle \frac{0.001\times 6.02\times {10}^{23}}{232}}$

Decay rate per second $={\displaystyle \frac{22\pi }{15}}$

$$\begin{array}{rl}& \therefore -{\displaystyle \frac{22\pi }{15}}=-{\displaystyle \frac{\lambda \times 0.001\times 6.02\times {10}^{23}}{232}}\\ & \lambda ={\displaystyle \frac{232\times 22\pi }{15\times 0.001\times 6.023\times {10}^{23}}}=1.775\times {10}^{-12}\\ & {\mathrm{T}}_{1/2}={\displaystyle \frac{0.693}{\lambda }}-1.23\times {10}^{10}\text{ year }\end{array}$$

Hence option (4) is correct.

Difficult

Radioactivity

65. A does of $5\mathrm{mc}$ of ${}_{15}^{30}\mathrm{P}$ is administred intraenously to a patient whose blood volume is 3.5 litre. At the end of one hour it is assumed that the phosporous is uniformly distributed. Then the count rate after per $\mathrm{ml}$ of withdrawn blood ofter one hour is (assume that the counter had an efficiency of only $10\mathrm{\%}$ ). Given $1\mathrm{mc}=3.7\times {10}^7\mathrm{dis}/\mathrm{s}$ and half life of phosphorous is 14 day.

(1) $5.3\times {10}^2\mathrm{dis}/\mathrm{s}$

(2) $5.0\times {10}^3\mathrm{dis}/\mathrm{s}$

(3) $5.3\times {10}^7\mathrm{dis}/\mathrm{s}$

(4) $530\mathrm{dis}/\mathrm{s}$

Show Answer

Correct answer: (1)

Solution:

Dose per $\mathrm{ml}={\displaystyle \frac{5}{3500}}\mathrm{mc}$

Number of disintegration ${\displaystyle \frac{1}{700}}\times 3.7\times {10}^7$ per sec. in blood

After one hour we can assume no decay.

Disintegration counted $={\displaystyle \frac{1}{10}}\times {\displaystyle \frac{1}{700}}\times 3.7\times {10}^7=5.3\times {10}^2\mathrm{dis}/\mathrm{s}$

Hence option(1) is correct.

Average

Radioactivity

66. Masses of two isobars ${}_{29}{\mathrm{Cu}}^{64}$ and ${}_{30}{\mathrm{Zn}}^{64}$ are $63.9298\mathrm{u}$ and $63.9292\mathrm{u}$. It can be concluded from this

(1) Both isobars are stable (Incorrect)

(2) ${\mathrm{Zn}}^{64}$ is radioactive decaying to ${\mathrm{Cu}}^{64}$ through $\beta$ decay

(3) ${\mathrm{Cu}}^{64}$ is radioactive decaying to ${\mathrm{Zn}}^{64}$ through $\gamma$ decay

(4) ${\mathrm{Cu}}^{64}$ is radioactive decaying to ${\mathrm{Zn}}^{64}$ through $\beta$ decay

Show Answer

Correct answer: (4)

Solution:

In $\beta$ decay process atomic number increases by 1 whereas mass number remains same. Hence the possible nuclear reaction is

$${}_{29}{\mathrm{Cu}}^{64}\to {}_{30}{\mathrm{Zn}}^{64}+{}_{-1}{\mathrm{e}}^0$$

Hence option (4) is correct.

Difficult

Radioactivity

67. A radioactive sample consist of two distinct spices having equal number of atoms intially. The mean life one spices is $\tau$ and that ofother $5\tau$. The decay products in both cases are stable. The following figure shows graphs $P,Q,R,S$ the plot of the total number of radioactive nuclei as a function of time. Which of the graphs $P,Q,R$ and $S$ represent the best form of the plot?

(1) $\mathrm{P}$

(2) $\mathrm{Q}$

(3) $\mathrm{R}$

(4) $\mathrm{S}$

Show Answer

Correct answer: (4)

Solution:

Since it is the process of radioactivity decay the total number of atoms cannot remains constant (as in option 1). Also the total number of atoms cannot ever increase (as in option 2 and 3). The total number of atoms can only decrease with time. Hence option (4) is the best plot.

Difficult

Radioactivity

68. An accident occurs in a laboratory in which a large amount of radioactive material with a known half life of 30 day becomes embedded on the floor walls etc. Tests show that the level of radiation is 32 times the permissible level of normal occupancy of the room. Assuming that the last statement is corerct, after how many days the laboratory can be safety occupied after.

(1) 32 days

(2) 100 days

(3) 20 days

(4) Lab can safety be reoccupied after an infinite time

Show Answer

Correct answer: (2)

Solution:

Since the initial value of radiation is 32 times the permissible level we have to find $t$ after which the activity drops to $1/32$ of its initial value which implies that we must have

$${\displaystyle \frac{\mathrm{N}}{{\mathrm{N}}_0}}={\displaystyle \frac{1}{32}}={\displaystyle \frac{1}{(2{)}^5}}$$

Since the half life is 20 days, the lab will be safe for use after 5 half-lives i.e., $\mathrm{t}=20\times 5=100$ days

Hence option (2) is correct.

Difficult

Radioactivity

69. $10\mathrm{mg}$ of carbon from living material produce 200 counts per minute due to a small proportion of radioactive isotope carbon 14. A piece of ancient wood of mass $10\mathrm{mg}$ is found to give 50 counts per minute. Assuming that carbon 14 content of the atomsphere has remained unchanged the estimated age of the wood sample is (Given half life of carbon is 5700 years)

(1) $1.41\times {10}^7$ years

(2) $1.41\times {10}^{10}$ years

(3) $1.41\times {10}^4$ years

(4) $1.41\times {10}^{14}$ years

Show Answer

Correct answer: (3)

Solution:

Since the amount of carbon-14 in atomsphere remain constant and a living matter acquires carbon from atmosphere, the count from a living matter represent the amount of carbon-14 present in atmosphere. The ancient wood which has now 50 counts per minute from $10\mathrm{mg}$ in the begining must have the same count rate as from the living matter that is 200 count per minute from $10\mathrm{mg}$. Thus the age of the ancient wood is the time in which the carbon-14 count has decreased from the initial value of 200 counts per minute from $10\mathrm{mg}$ to the find value of 50 counts per minute from $10\mathrm{mg}$. Then $\lambda$ the disintegration constants of ${\mathrm{C}}^{14}$ is given by

$$\lambda ={\displaystyle \frac{0.693}{5700}}{\text{ year }}^{-1}(\because {\mathrm{T}}_{1/2}=5700\text{ years })$$

Representing by $t$ the age of ancient wood we have

$$\begin{array}{rl}& 50=200e^{-\left({\displaystyle \frac{0.693}{5700}}\right)\mathrm{t}}\text{ or }4={\mathrm{e}}^{{\displaystyle \frac{0.693}{5700}}\mathrm{t}}\\ & \mathrm{t}={\displaystyle \frac{2.302\times 6021\times 5700}{0.693}}=1.41\times {10}^4\text{ years }\end{array}$$

Hence option (3) is coorect.

Easy

Nucleus

70. The total kinetic energy, of the electron antineutrinos pair, in the $\beta$-decay process, equals $8.10\times {10}^6\mathrm{eV}$. The antineutrinos, formed in this decay, could escape with a maximum kinetic energy of

(1) zero

(2) Much less than $8.0\times {10}^6\mathrm{eV}$

(3) Nearly $0.8\times {10}^6\mathrm{eV}$

(4) Much larger than $0.8\times {10}^6\mathrm{eV}$

Show Answer

Correct answer: (2)

Solution:

The following reaction represents $\beta$ decay process

$$\mathrm{n}\to \mathrm{p}+\bar{\mathrm{e}}+\bar{\mathrm{v}}$$

In this decay process the kinetic energy of the pair of electron and antineutrons is constant. When K.E of electron is maximum, K.E. of $v$ is zero. The antineutrinos will have maximum energy when K.E of electron is zero.

Maximum K.E of antineutron will be nearly $0.8\times {10}^6\mathrm{eV}$.

Hence option (3) is correct.

Difficult

Bohr’s Theory

71. Out of the following which one is not a possible energy for a photon to be emitted by a hydrogen atom.

(1) $1.9\mathrm{eV}$

(2) $11.1\mathrm{eV}$

(3) $13.6\mathrm{eV}$

(4) $0.65\mathrm{eV}$

Show Answer

Correct answer: (2)

Solution:

According to Bohr’s theory the energy of a photon emitted when it makes a transition from a quantum state ${\mathrm{n}}_2$ to ${\mathrm{n}}_1$ is given by

$$\mathrm{E}=\mathrm{h}v={\mathrm{E}}_{{\mathrm{n}}_2}-{\mathrm{E}}_{{\mathrm{n}}_1}=\mathrm{R}({\displaystyle \frac{1}{{\mathrm{n}}_1^2}}-{\displaystyle \frac{1}{{\mathrm{n}}_2^2}})$$

Also the value of $E$ corresponding to quantum number $n$ is given by $E_n=-{\displaystyle \frac{13.6}{n^2}}eV$

From above we get

$$\begin{array}{ll}{\mathrm{E}}_1=-13.6\mathrm{eV}& {\mathrm{E}}_2=-3.4\mathrm{eV}\\ {\mathrm{E}}_3=0.85\mathrm{eV}& {\mathrm{E}}_4=0.58\mathrm{eV}\end{array}$$

It is clear that $11.1\mathrm{eV}$ is not obtainable for the difference of any one of the above given values.

Hence option (2) is correct.

Easy

Nucleus

72. During a negative beta decay

(1) an atomic electron is ejected

(2) an electron which is already present within the nucleus is ejected

(3) A neutron in the nucleus decays emitting an electron

(4) A part of the binding energy of the nucleus is converted into an electron

Show Answer

Correct answer: (3)

Solution:

During a negative beta decay, the reaction which takes place is

$${}_0^1\mathrm{n}\to {}_1^1\mathrm{p}+{}_{-1}^0\mathrm{e}+\overline{v}$$

Here $\overline{v}$ is an anti neutrinos

Hence option(3) is correct.

Difficult

Binding Energy

73. In the photo disintegration equation for the deutron the reaction energy is $2.227\pm \mathbf{0}\mathbf{.}\mathbf{0}\mathbf{0}\mathbf{3}$ $\mathrm{MeV}$ is

$$\gamma (h\nu )+{}_1{\mathrm{H}}^2\to {}_1{\mathrm{H}}^2+{}_0{n}^1+2.27\mathrm{MeV}$$

From the mass spectrometer the mass differences between the doublet $2\left({}_1{\mathrm{H}}^2\right)$ and $\mathrm{D}\left({}_1{\mathrm{H}}^2\right)$ is $1.5380\pm 0.0021\times {10}^{-2}\mathrm{amu}$. If the mass of hydrogen atom is $1.008145\pm 0.000003\mathrm{amu}$. Then the mass of neutron is given by

(1) $1.0089992+0.0000021\mathrm{amu}$

(2) $1.0089992+0.000003\mathrm{amu}$

(3) $1.0089992+0.000008\mathrm{amu}$

(4) $1.0089992+0.003\mathrm{amu}$

Show Answer

Correct answer: (2)

Solution:

The mass of neutron is given by

$$\mathrm{M}\left({}_1{\mathrm{H}}^2\right)-\mathrm{M}\left({}_1{\mathrm{H}}^1\right)-{\displaystyle \frac{2.27}{931}}\mathrm{amu}$$

$$=-0015380+1.008145-0.002392=1.0089999\mathrm{amu}$$

Errors:

(i) Energy $2.27\pm 0.003\mathrm{MeV}$ equivalent of an error of 0.000003

(ii) Doublet error $=\pm 0.0000021\mathrm{amu}$

(iii) Hydrogen atom $=0.00003\mathrm{amu}$

Total maximum error is $\pm 0.000008\mathrm{amu}$

Hence the mass of neutron is $1.0089992\pm 0.000008\mathrm{amu}$

Hence option(3) is correct.

Difficult

Binding Energy

74. The binding energy per nucleon for deuteron and helium are $1.1\mathrm{MeV}$ and $7.0\mathrm{MeV}$ respectively. The two deutrons fuse to form a helium nucleus. The energy released in this fusion reaction is

(1) $23.6\mathrm{MeV}$

(2) $32.6\mathrm{MeV}$

(3) $9.2\mathrm{MeV}$

(4) $26.9\mathrm{MeV}$

Show Answer

Correct answer: (1)

Solution:

Binding energy per nucleon for deutron $=1.1\mathrm{MeV}$

Total binding energy of deutron $=4\times 1.1=4.4\mathrm{MeV}$

Binding energy per nucleon for helium $=7.0\mathrm{MeV}$

Total binding of helium $14\times 7=28.0\mathrm{MeV}$

Energy released $=28-4.4=23.6\mathrm{MeV}$

Hence option (1) is correct.

Difficult

Binding Energy

75. The curve in the figure shows the binding energy per nucleon versus mass number for four nuclei $\mathrm{W},\mathrm{X},\mathrm{Y}$ and $\mathrm{Z}$. The process that would release energy is

(1) $\mathrm{Y}\to 2\mathrm{Z}$

(2) $\mathrm{W}\to \mathrm{X}+\mathrm{Z}$

(3) $\mathrm{W}\to 2\mathrm{Y}$

(4) $\mathrm{X}\to \mathrm{Y}+\mathrm{Z}$

Show Answer

Correct answer: (3)

Solution:

We know that energy is released when the total binding energy of products is more than the rectants.

In option (3) we find that.

Binding energy of reactants $=120\times 7.5=900\mathrm{MeV}$

Binding energy of products $=260\times 8.5=1020\mathrm{MeV}$

Thus $1020\mathrm{MeV}>900\mathrm{MeV}$ and this process would release energy.

In option 1 B. . of reactants $=60\times 7.5=450$

B.E of products $=2(30\times 50)=300$

Not possible

Similar is the case for process (2) and (4).

Hence option(3) is correct.

Average

Nuclear Nuclear Reactions

76. ${}_{10}^{22}\mathrm{Ne}$ nucleus after absorbing energy decays into two $\alpha$ particles and an unknown nucleus. The unknown nucleus is

(1) nitrogen

(2) carbon

(3) boron

(4) oxygen

Show Answer

Correct answer: (2)

Solution:

In a nuclear reaction, the mass number and atomic number of reactants and products are conserved.

Hence

$$\begin{array}{rl}& {\mathrm{Ne}}_{10}^{22}+\text{ energy }\to 2{\mathrm{He}}_2^4+\mathrm{X}\\ & {\mathrm{Ne}}_{10}^{22}\to 2{\mathrm{He}}_2^4+{\mathrm{X}}_6^{14}\end{array}$$

Obvisouly $\mathrm{X}$ is carbon with mass number 14 and atomic number 6 .

Hence option (2) is correct.

Average

Nuclear Reactions

77. The nucleus of ${\mathbf{P}\mathbf{u}}^{238}$ decays to $U^{234}$ by emitting a $\alpha$ particle. The kinetic energy of the emitted $\alpha$ particle is given by [you are given that atomic mass of ${\mathrm{Pu}}^{238}$ is $238.04954\mathrm{amu}$ that of ${\mathrm{U}}^{234}$ is $234.040969\mathrm{amu}$ ].

(1) $0059\mathrm{MeV}$

(2) $5.57\mathrm{amu}$

(3) $2.76\mathrm{MeV}$

(4) $5.57\mathrm{MeV}$

Show Answer

Correct answer: (4)

Solution:

The reaction

$${\mathrm{P}}^{238}\to {\mathrm{U}}^{234}+{}_2{\mathrm{He}}^4+\mathrm{Q}$$

Represents the decay process of ${\mathrm{Pu}}^{238}$

Here $Q$ is representing the energy of nuclear reaction

$\mathrm{Q}=\mathrm{m}\left({\mathrm{Pu}}^{238}\right)-[\mathrm{m}\left({\mathrm{U}}^{234}\right)+\mathrm{m}\left({}_2{\mathrm{He}}^4\right)]$

$\mathrm{Q}=238.04954-[234.04096+4.002603]=0.00598\mathrm{amu}=0.00598\times 931=5.57\mathrm{MeV}$

This energy would have to shared by ${\mathrm{U}}^{234}$ and $\alpha$ particle. However ${\mathrm{U}}^{234}$ is very massive as compared to the mass of the $\alpha$ particles. Hence this energy of $5.57\mathrm{MeV}$ can be taken as the kinetic energy of the emitted $\alpha$ particle.

Hence option (4) is correct.

Difficult

Atoms and Nuclei

78. In the options given below, let E denotes the rest mass energy of a nucleus and n represent a neutron. The correct option is

(1) $\mathrm{E}\left({}_{92}^{236}\mathrm{U}\right)>\mathrm{E}\left({}_{53}^{137}\mathrm{I}\right)+\mathrm{E}\left({}_{39}^{97}\mathrm{Y}\right)+2\mathrm{E}(\mathrm{n})$

(2) $\mathrm{E}\left({}_{92}^{236}\mathrm{U}\right)<\mathrm{E}\left({}_{43}^{137}\mathrm{I}\right)+\mathrm{E}\left({}_{29}^{97}\mathrm{X}\right)+2\mathrm{E}(\mathrm{n})$

(3) $\mathrm{E}\left({}_{92}^{239}\mathrm{U}\right)<\mathrm{E}\left({}_{56}^{140}\mathrm{Ba}\right)+\mathrm{E}\left({}_{36}^{94}\mathrm{Kr}\right)+2\mathrm{E}(\mathrm{n})$

(4) $\mathrm{E}\left({}_{92}^{235}\mathrm{U}\right)<\mathrm{E}\left({}_{56}^{140}\mathrm{Ba}\right)+\mathrm{E}\left({}_{36}^{94}\mathrm{Kr}\right)+\mathrm{E}(\mathrm{n})$

Show Answer

Correct answer: (1)

Solution:

Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei. Hence option (1) is correct.

79. The fusion reaction $2_1{\mathrm{H}}^2{\to }_2{\mathrm{He}}^4+$ energy is proposed to use for the production of industrial power. Assuming the efficiency of the process to be $30\mathrm{\%}$, the number of deutrons atoms per second required to produce an output of $50000\mathrm{kW}$ is of the order of [Given mass of ${}_1{\mathrm{H}}^2=\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{1}\mathbf{4}\mathbf{7}\mathbf{8}\mathrm{amu}$ and mass of ${}_1{\mathrm{He}}^2=\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{0}\mathbf{3}\mathbf{8}\mathbf{8}\mathrm{amu}]$.

(1) $\sim {10}^{17}$

(2) $\sim {10}^{18}$

(3) $\sim {10}^{19}$

(4) $\sim {10}^{20}$

Show Answer

Correct answer: (4)

Solution:

The reaction is represented by $2{}_1{\mathrm{H}}^2\to {}_2{\mathrm{He}}^4+$ energy

Mass defect $\mathrm{\Delta }\mathrm{m}[2\times 2.04178-4.00388]=0.02568$ amuu

Energy released $=0.02568\times 931\mathrm{MeV}=23.91\mathrm{MeV}$

Since efficiency of the process is $30$ actual output is $23.91\times {\displaystyle \frac{30}{100}}=7.173\mathrm{MeV}$

Actual output per deutron atom is ${\displaystyle \frac{7.173}{2}}=3.587\mathrm{MeV}=3.587\times 1.6\times {10}^{-12}\mathrm{J}$

Output required $=50000\mathrm{kW}=5\times {10}^7\mathrm{J}=5\times {10}^7\mathrm{J}$

No. of deutron atomsrequired to produce this output

$$={\displaystyle \frac{5\times {10}^7}{3.587\times 1.6\times {10}^{-12}}}=871.4\times {10}^{17}\approx {10}^{20}\mathrm{atoms}/\mathrm{s}$$

Hence option (4) is correct.

Average

Nuclear Fusion

80. A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in fusion reaction is $0.02866\mathrm{u}$. The energy liberated per $u$ is $(1\mathrm{u}=931\mathrm{MeV})$

(1) $13.35\mathrm{MeV}$

(2) $2.67\mathrm{MeV}$

(3) $26.7\mathrm{MeV}$

(4) $6.67\mathrm{MeV}$

Show Answer

Correct answer: (4)

Solution:

Total mass defect $\mathrm{\Delta }\mathrm{m}$ in the fusion reaction is $0.02866\mathrm{u}$

Total energy released $=0.02866\times 931\mathrm{MeV}$

Since 4 nuclei of hydrogen fuse to form a helium nucleus.

$\therefore$ Energy realeased per nucleus $={\displaystyle \frac{0.02866\times 931}{4}}=6.67\mathrm{MeV}$

Hence option (4) is correct.

Difficult

Nuclear Fusion

81. A star initially has ${10}^{40}$ deutrons. It produces energy via the process ${}_1{\mathbf{H}}^2+{}_1{\mathbf{H}}^2+p$ and ${}_1{\mathrm{H}}^2+{}_1{\mathrm{H}}^3{\to }_2{\mathrm{He}}^4+\mathrm{n}$. If the average power radiated by the star is ${10}^{16}\mathrm{W}$, the deutron supply of the star is exhaused in a time of the order of

(1) ${10}^6\mathrm{s}$

(2) ${10}^8\mathrm{s}$

(3) ${10}^{12}\mathrm{s}$

(4) ${10}^{16}\mathrm{s}$

Show Answer

Correct answer: (3)

Solution:

The given reactions are

$$\begin{array}{rl}& {}_1^2\mathrm{H}+{}_1^2\mathrm{H}\to {}_1^3\mathrm{H}+\mathrm{p}\\ & {}_1^2\mathrm{H}+{}_1^3\mathrm{H}\to {}_2^4\mathrm{He}+\mathrm{n}\\ & 3\left({}_1^2\mathrm{H}\right)+{}_2^4\mathrm{He}+\mathrm{n}+\mathrm{p}\end{array}$$

Mass defect $\mathrm{\Delta }\mathrm{m}=(3\times 2.014-4.001-1.007-1.0008)=0.026\mathrm{amu}$

Energy released $=0.026\times 931\mathrm{MeV}$

$$=0.026\times 931\times 1.6\times {10}^{-13}=3.87\times {10}^{-13}\mathrm{J}$$

This is the energy produced by the consumption of 3 deutron atoms.

The energy released by ${10}^{40}$ deutrons $={\displaystyle \frac{{10}^{40}\times 3.87\times {10}^{-12}}{3}}=1.29\times {10}^{28}\mathrm{J}$

The average power radiated is ${10}^{16}\mathrm{W}={10}^{16}\mathrm{J}/\mathrm{s}$

$\therefore$ Time required to exhaust all the deutron of the star will be given by ${\displaystyle \frac{1.29\times {10}^{28}}{{10}^{16}}}=1.29\times {10}^{12}\mathrm{s}\approx {10}^{12}$

Hence option (3) is correct.