Solution:

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

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

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

$$ r _{1}=r _{n}^{\prime} $$

$\therefore$ we must have

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

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

Solution:

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

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

$$ =\mathrm{z}^{2}\left(-\dfrac{13.6}{\mathrm{n}^{2}}\right)=-\dfrac{4 \times 13.6}{\mathrm{n}^{2}}=-\dfrac{54.4}{\mathrm{n}^{2}} \mathrm{eV} $$

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 $\mathrm{n}=2$

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

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 $\quad \mathrm{V} _{1}=\dfrac{40.8}{\mathrm{e}} \mathrm{eV}=40.8 \mathrm{Volt}$

Solution:

For Lyman series we have

$$ \bar{\nu}=\dfrac{1}{\lambda}=\mathrm{R}\left(\dfrac{1}{1^{2}}-\dfrac{1}{\mathrm{n}^{2}}\right) $$

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

Hence $\dfrac{1}{\lambda _{\mathrm{S}}}=\mathrm{R}\left(\dfrac{1}{1}-\dfrac{1}{\infty}\right)=\mathrm{R}$

$$ \lambda _{\mathrm{s}}=\dfrac{1}{\mathrm{R}}=\dfrac{1}{10967700} \mathrm{~m}=911.6 \AA $$

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

$\therefore \dfrac{1}{\lambda _{\mathrm{L}}}=\mathrm{R}\left(\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right)=\dfrac{3}{4} \mathrm{R}$

or

$$ \lambda _{\mathrm{L}}=\dfrac{4}{3} \times \dfrac{1}{\mathrm{R}}=\dfrac{4}{3} \dfrac{1}{10967700} \mathrm{~m}=1215 \AA $$

Solution:

The Possible states when transition occur from $\mathrm{n}^{\text {th }}$ state are $(\mathrm{n}-1)^{\text {th }} \ldots . . .2^{\text {nd }}$ 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

$$ \begin{aligned} & (\mathrm{n}-1)+(\mathrm{n}-2)+(\mathrm{n}-3)+\ldots \ldots \ldots \ldots+2+1\\ & =\dfrac{\mathrm{n}(\mathrm{n}-1)}{2} \end{aligned} $$

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.

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

Solution:

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

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

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

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

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

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

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

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

Solution:

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

Half life of radon $=3.825$ days

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

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

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

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 $\quad \mathrm{e}^{5 \lambda}=\dfrac{4750}{2700}=1.76$

$\lambda=\dfrac{0.5654}{5}=0.1131 \mathrm{~m}^{-1}$

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

Solution:

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

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

$$ \lambda=\dfrac{0.693}{26.8 \times 60} \mathrm{~s}^{-1} $$

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

$$ \begin{gathered} \dfrac{\mathrm{dN}}{\mathrm{dt}}=\lambda \mathrm{N}\\ \mathrm{N}=\dfrac{3.7 \times 10^{10} \times 26.8 \times 60}{0.693} \end{gathered} $$

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

Weight of RaB having an activity of 1 curie

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

Solution:

$$ \text { Half life } \mathrm{T}=\dfrac{0.693}{\lambda}=\dfrac{0.693}{0.00231}=300 \mathrm{~s} \quad\left[\lambda=0.00231 \overline{\mathrm{s}}^{1}\right] $$

Mean life $\zeta=\dfrac{1}{\lambda}=\dfrac{1}{0.00231}=432.95 \mathrm{~s}$

Solution:

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

Kinetic energy available in the given reaction is

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

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

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.

$$ \begin{gathered} \mathrm{A} _{1}=\dfrac{1}{3}(235)\\ \therefore \quad \mathrm{r} _{1}=(1.4 \mathrm{fm})\left(\dfrac{235}{3}\right)^{1 / 3}=5.99 \mathrm{fm}\\ \text { and } \quad \mathrm{A} _{2}=\dfrac{2}{3}(235)\\ \therefore \quad \mathrm{r} _{2}=(1.4 \mathrm{fm})\left(\dfrac{2}{3 \times 235}\right)^{1 / 3}=7.55 \mathrm{fm} \end{gathered} $$

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 $\dfrac{1}{2} \mathrm{mv}^{2}=0.0327 \mathrm{eV}=0.0327 \times 1.6 \times 10^{-19} \mathrm{~J}$

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

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

$$ \mathrm{dt}=\dfrac{10}{2.5 \times 10^{3}}=4 \times 10^{-3} \mathrm{~s} $$

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

$$ \begin{aligned} & \dfrac{\mathrm{dN}}{\mathrm{N}}=\lambda \mathrm{dt}\\ & \text { or } \quad \dfrac{\mathrm{dN}}{\mathrm{N}}=\dfrac{0.693}{700} \times 4 \times 10^{-3}\\ & =8.96 \times 10^{-4} \end{aligned} $$

Solution:

The reaction can be written as

$$ { } _{1} \mathrm{H}^{2}+{ } _{1} \mathrm{H}^{2} \rightarrow{ } _{2} \mathrm{He}^{4}+\text { Energy } $$

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

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

Solution:

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

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

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

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 $\dfrac{1}{4 \pi \varepsilon _{0}} \dfrac{47 \times 47}{\left(11.77 \times 10^{-15}\right)^{2}}$

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

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 $=\dfrac{10^{6}}{3.2 \times 10^{-11}}=3.125 \times 10^{6}$

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 $=\dfrac{9.9 \times 10^{23} \times 235}{6.02 \times 10^{23}}$

$$ =386.46 \mathrm{~g} $$

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 $\dfrac{6.64 \times 10^{-27}}{3.2 \times 10^{-12}} \times 8640000 \times 10^{6}$

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

Solution:

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

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

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

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

Since the abudance of deutorium is $0.0156 \%$, 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

$$ \dfrac{43}{6}=7.17 \mathrm{MeV} $$

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?

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 $90^{\circ}$.

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

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.

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

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?

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

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.

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

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.

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.

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

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.

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.

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.

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.

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

$$ { } _{94} \mathrm{P}^{239} \rightarrow{ } _{92} \mathrm{U}^{235}+{ } _{2} \mathrm{He}^{4} $$

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

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.

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.

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

22. Find the mass of ${ } _{6}^{14} \mathrm{C}$ in the following reaction

$$ { } _{7}^{14} \mathrm{~N}+{ } _{0} \mathrm{n}^{1} \rightarrow{ } _{6}^{14} \mathrm{C}+{ } _{1}^{1} \mathrm{H}+0.55 \mathrm{MeV} $$

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.

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

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

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.

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 ${ } _{3}^{7} \mathrm{Li}=7.01599 \mathrm{amu}$.

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 $\mathrm{B}^{11}$.

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

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.

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?

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 $4 \mathrm{~kW}$.

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 $20 \%$ efficient. How many uranium atoms does it consume in one day? What mass of uranium does it consume each day?

32. Calculate the energy released in the fusion reaction

$$ { } _{2}^{4} \mathrm{He}+{ } _{2}^{4} \mathrm{He}+{ } _{2}^{4} \mathrm{He} \rightarrow{ } _{6}^{12} \mathrm{C} $$

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

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 $=4.00263 \mathrm{amu}$

34. It is proposed to use nuclear fusion reaction

$$ { } _{1}^{2} \mathrm{H}+{ } _{1}^{2} \mathrm{H} \rightarrow{ } _{2}^{4} \mathrm{He} $$

in a nuclear reactor of $200 \mathrm{MW}$ rating. If the energy from above reaction is used with a $25 \%$ 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.

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

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

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

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?

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.

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 $\dfrac{\mathrm{h}}{2 \pi}$.

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 $\mathrm{E} _{\mathrm{n}}=-\dfrac{13.6}{\mathrm{n}^{2}} \mathrm{eV}$

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: $\mathrm{T}=\dfrac{0.693}{\lambda}$.

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. $\tau=\dfrac{1}{\lambda}$.

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

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) $\lambda _{1}=2 \lambda _{2}=3 \lambda _{3}=4 \lambda _{4}$

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 $1.4 \mathrm{eV}$

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 $V=V _{0} \ell n\left(\dfrac{r}{r _{0}}\right)$

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

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

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

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

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

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) $\sqrt{\dfrac{3}{5 \alpha}}$

(2) $\sqrt{\dfrac{9}{2 \alpha}}$

(3) $\sqrt{\dfrac{27}{5 \alpha}}$

(4) $\sqrt{\dfrac{5}{27 \alpha}}$

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 $(\mathrm{n} \neq 0)$ is

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

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

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

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

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 $=16 u)$, 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 $\left(4 / \pi \times 10^{11}\right) \mathbf{H z}$ (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 $1.17 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^{2}$

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) $\dfrac{9 \mathrm{~h}^{2} \mathrm{R}^{2}}{256 \mathrm{M}}$

(2) $\dfrac{9 \mathrm{~h}^{2} \mathrm{R}^{2}}{512 \mathrm{M}}$

(3) $\dfrac{25 \mathrm{~h}^{2} \mathrm{R}^{2}}{1296 \mathrm{M}}$

(4) $\dfrac{25 \mathrm{~h}^{2} \mathrm{R}^{2}}{512 \mathrm{M}}$

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

$$ \dfrac{1}{\lambda}=\mathrm{R}\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(4)^{2}}\right]=\dfrac{3 \mathrm{R}}{16} $$

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

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 $=\dfrac{\mathrm{p}^{2}}{2 M}=\dfrac{9 \mathrm{~h}^{2} \mathrm{R}^{2}}{512 \mathrm{M}}$

Hence option (2) is correct.

Correct answer: (4)

Solution:

We have $\mathrm{w} _{0}=$ work function of photosensitive material $=\dfrac{\mathrm{hc}}{\lambda _{0}}$. Let $\mathrm{E}=\mathrm{h} v$, be the energy of photon incident on the photosensitive material. FromEinstein’s equation of photoelectric effect, we have

$2 \mathrm{w} _{0}=\mathrm{E}-\mathrm{w} _{0} \quad \therefore \quad \mathrm{E}=3 \mathrm{w} _{0}=\dfrac{3 \mathrm{hc}}{\lambda _{0}}$ $\hspace{5 cm}$ …….(1)

From Bohr’s theory, $\quad \mathrm{E}=\left(\mathrm{E} _{\mathrm{m}+1}-\mathrm{E} _{1}\right)=(\alpha)\left[1-\dfrac{1}{(\mathrm{~m}+1)^{2}}\right]$ $\hspace{2 cm}$ …….(2)

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

$$ \begin{aligned} & \dfrac{3 \mathrm{hc}}{\lambda _{0}}=\alpha\left[1-\dfrac{1}{(\mathrm{~m}+1)^{2}}\right] \quad \text { or } \quad \dfrac{1}{(\mathrm{~m}+1)^{2}}=\left[1-\dfrac{3 \mathrm{hc}}{\alpha \lambda _{0}}\right]\\ \therefore \mathrm{m} & =\left[\dfrac{\alpha \lambda _{0}}{\alpha \lambda _{0}-3 \mathrm{hc}}\right]^{1 / 2}-1 \end{aligned} $$

Hence option (4) is correct.

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

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

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

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

$\therefore E _{n}=\dfrac{n^{2} h^{2}}{8 m^{2}}$

For a given value of m; $E _{n} \propto a^{-2}$

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

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

Hence option(1) is correct.

Correct answer: (2)

Solution:

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

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

$\therefore 4.5=\dfrac{(3)^{2}}{\mathrm{z}} \quad \therefore \mathrm{z}=2$

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

$$ \dfrac{1}{\lambda}=\mathrm{Rz}^{2}\left[\dfrac{1}{\mathrm{n} _{1}^{2}}-\dfrac{1}{\mathrm{n} _{2}^{2}}\right] $$

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

(i) $\mathrm{n} _{2}=3 \rightarrow \mathrm{n} _{1}=1$; (ii) $\mathrm{n} _{2}=3 \rightarrow \mathrm{n} _{1}=2$; and (iii) $\mathrm{n} _{2}=2 \rightarrow \mathrm{n} _{1}=1$.

The corresponding wavelengths are

$$ \begin{aligned} \dfrac{1}{\lambda _{1}} & =(2)^{2} \mathrm{R}\left[\dfrac{1}{1}-\dfrac{1}{9}\right]=\dfrac{32 \mathrm{R}}{9} & & \therefore \lambda _{1}=\dfrac{9}{32 \mathrm{R}}\\ \dfrac{1}{\lambda _{2}} & =(2)^{2} \mathrm{R}\left[\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right]=\dfrac{5 \mathrm{R}}{9} & & \therefore \lambda _{2}=\dfrac{9}{5 \mathrm{R}}\\ \text { and } \dfrac{1}{\lambda _{3}} & =(2)^{2} \mathrm{R}\left[\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right]=3 \mathrm{R} & & \therefore \lambda _{3}=\dfrac{1}{3 \mathrm{R}} \end{aligned} $$

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

Hence option (2) is correct.

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 $\mathrm{E} _{2 \mathrm{n}}=\left[\dfrac{\mathrm{E} _{1}}{(2 \mathrm{n})^{2}}\right]$. The maximum energy photon is emitted when electron jumps from this state to the ground state, i.e.,

$\quad\left(\dfrac{E _{1}}{4 \mathrm{n}^{2}}-E _{1}\right) =204 \mathrm{eV}$

$\therefore \quad E _{1}\left(\dfrac{1}{4 \mathrm{n}^{2}}-1\right) =204$ $\hspace{4 cm}$ …….(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

$\mathrm{E} _{1}\left[\dfrac{1}{4 \mathrm{n}^{2}}-\dfrac{1}{\mathrm{n}^{2}}\right]=40.8 \text { or }=\dfrac{3 \mathrm{E} _{1}}{4 \mathrm{n}^{2}}=40.8$ $\hspace{2 cm}$ …….(2)

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

$$ \begin{equation*} \dfrac{\left(1-\dfrac{1}{4 \mathrm{n}^{2}}\right)}{\left(\dfrac{3}{4 \mathrm{n}^{2}}\right)}=\dfrac{204}{40.8}=5 \tag{3} \end{equation*} $$

$\therefore \mathrm{n}=2$

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

$$ \begin{equation*} E _{n}=-\dfrac{4 \times(2)^{2} \times 40.8}{3} \mathrm{eV}=-217.6 \mathrm{eV} \tag{4} \end{equation*} $$

But $\quad \mathrm{E} _{1}=\mathrm{E} _{1}$ (hydrogen) $\mathrm{z}^{2}=\left[-(13.6) \mathrm{z}^{2}\right] \mathrm{eV}$

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

$\therefore \mathrm{z}=4$

Hence option(1) is correct.

Correct answer: (3)

Solution:

For hydrogen atom; $E _{n}=\dfrac{-13.6}{n^{2}} \mathrm{eV}$

Hence $\mathrm{E} _{2}=\dfrac{-13.6}{4} \mathrm{eV}=-3.4 \mathrm{eV}$

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

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

$\therefore$ 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

$$ \begin{aligned} & \quad-3.4=-\dfrac{54.4}{\mathrm{n}^{2}}\\ & \therefore \quad \mathrm{n}^{2}=\dfrac{54.4}{3.4}=16\\ & \therefore \quad \mathrm{n}=4 \end{aligned} $$

Hence option (3) is correct.

Correct answer: (2)

Solution:

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

When this atom goes, from its first excited to its second excited state; $n _{1}=2$ and $n _{2}=3$. We are given that

$$ \begin{gathered} \mathrm{E} _{3}-\mathrm{E} _{2}=47.2 \mathrm{eV}\\ \therefore \quad\left(\mathrm{z}^{2}\right)(13.6)\left[\dfrac{1}{(2)^{2}}-\dfrac{1}{(3)^{2}}\right]=47.2 \end{gathered} $$

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

$\therefore \quad \mathrm{z}=5$

$\therefore \quad$ For the given atom,

$$ E _{n}=-\dfrac{13.6(5)^{2}}{n^{2}} e V=-\dfrac{340}{n^{2}} e V $$

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.

Correct answer: (1)

Solution:

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

$\mathrm{F}=-\dfrac{\mathrm{dU}}{\mathrm{dr}}=\dfrac{\mathrm{ke}^{2}}{\mathrm{r}^{4}}$ $\hspace{3 cm}$ …….(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

$\dfrac{m v _{n}^{2}}{r _{n}}=F=\dfrac{\mathrm{ke}^{2}}{\mathrm{r}^{4}}$ $\hspace{3 cm}$ …….(2)

From Bohr’s first postulate $\mathrm{mv} _{\mathrm{n}} \mathrm{r} _{\mathrm{n}}=\dfrac{\mathrm{nh}}{2 \pi}$ $\hspace{1 cm}$ …….(3)

From Eqn. (2) and (3), we get $v _{n}=\dfrac{n^{3} h^{3}}{8 \pi^{3} \mathrm{~km}^{2} e^{2}}$

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 $\mathrm{E} _{\mathrm{n}}=-\mathrm{K} _{\mathrm{n}}=-\dfrac{1}{2} \mathrm{mv} _{\mathrm{n}}^{2}$

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

Hence option(1) is correct.

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

$\mathrm{R} _{1}=\lambda _{1} \mathrm{~N} _{1} \quad \text { and } \quad \mathrm{R} _{2}=\lambda _{2} \mathrm{~N} _{2}$

$\therefore \dfrac{\mathrm{R} _{1}}{\mathrm{R} _{2}}=\left(\dfrac{\lambda _{1}}{\lambda _{2}}\right)\left(\dfrac{\mathrm{N} _{1}}{\mathrm{~N} _{2}}\right)$

$\therefore \dfrac{1}{3}=\left(\dfrac{\lambda _{1}}{\lambda _{2}}\right) \times 2$

$\therefore \dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{1}{6}$ $\hspace{3 cm}$ ……(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

$$ \dfrac{\tau _{2}}{\tau _{1}}=\dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{1}{6} $$

Given $\tau _{1}=1$ month. Therefore, $\tau _{2}=\dfrac{1}{6}$ month

Hence option(4) is correct.

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

$$ \mathrm{N} _{1}=\mathrm{N} _{0} \mathrm{e}^{-10 \lambda t} \text { and } \mathrm{N} _{2}=\mathrm{N} _{0} \mathrm{e}^{\lambda \mathrm{t}} $$

$\therefore \dfrac{\mathrm{N} _{1}}{\mathrm{~N} _{2}}=\mathrm{e}^{-9 \lambda \mathrm{T}}=\dfrac{1}{\mathrm{e}}=\mathrm{e}^{-1}$

$\therefore 9 \lambda t=1$

$\therefore \mathrm{t}=\dfrac{1}{9 \lambda}$

Hence option (4) is correct

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 $=\left|\dfrac{\mathrm{dN}}{\mathrm{dt}}\right|=\lambda \mathrm{N}$

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

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

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

Correct answer: (2)

Solution:

Given, $\tau _{1 / 2}=$ Half-life $=1$ day $=\dfrac{0.693}{\lambda}$ 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, $\Delta \mathrm{N}$, that have disintegrated in time interval $\left(\mathrm{t} _{2}-\mathrm{t} _{1}\right)$, is

$$ \begin{aligned} \Delta \mathrm{N} & =\left(\mathrm{N} _{2}-\mathrm{N} _{1}\right)=\mathrm{N} _{0}\left(\mathrm{e}^{-3}-\mathrm{e}^{-2}\right)\\ \therefore \quad \dfrac{\Delta \mathrm{N}}{\mathrm{N} _{0}} & =\mathrm{e}^{-3}-\mathrm{e}^{-2}=\left(\dfrac{1-\mathrm{e}}{\mathrm{e}^{3}}\right)=\left(\dfrac{-\mathrm{e}+1}{\mathrm{e}^{3}}\right) \end{aligned} $$

Hence option (2) is correct.

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

$$ \mathrm{X}=\mathrm{N} _{0}\left[1-\mathrm{e}^{-\lambda \tau}\right]=\mathrm{N} _{0}\left[1-\mathrm{e}^{-1}\right] \hspace {3 cm}\left(\because \tau=\dfrac{1}{\lambda}, \text { where } \lambda=\text { decay constant }\right) $$

and $\quad \mathrm{Y}=\mathrm{N} _{0}\left[1-\mathrm{e}^{-3 \lambda \bar{\tau}}\right]=\mathrm{N} _{0}\left[1-\mathrm{e}^{-3}\right]$

Percentage change in counts in the time interval $\left(\mathrm{t} _{1}=\tau\right)$ to $\left(\mathrm{t} _{2}=3 \tau\right)$, is $=\left(\dfrac{\mathrm{Y}-\mathrm{X}}{\mathrm{X}}\right) \times 100 \%$

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

Hence option (2) is correct.

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 $\left(N=N _{1}\right)$, at $\left(t _{1}=\tau=10\right.$ day $)$, we have

$\quad \mathrm{N} _{1}=\mathrm{N} _{0}\left(\mathrm{e}^{-1}\right)$ $\hspace{5 cm}$……(1)

At $t _{2}$ (= 30 days), the total number of atoms, that have disintegrated, is

$\quad \mathrm{N} _{2}=\mathrm{N} _{0}\left[1-\mathrm{e}^{-3}\right]$ $\hspace{5 cm}$……(2)

$\therefore \quad \dfrac{\mathrm{N} _{1}}{\mathrm{~N} _{2}}=\dfrac{\mathrm{e}^{-1}}{1-\mathrm{e}^{-3}}=\dfrac{\mathrm{e}^{2}}{\mathrm{e}^{3}-1}$

Hence option(3) is correct.

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

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

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

$$ \begin{aligned} & A _{1}=A^{\prime}=(A-4 x)\\ & z _{1}=\left(z^{\prime}-y\right)=(z-2 x-y) \end{aligned} $$

Hence option (2) is correct.

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 $=\left(\dfrac{\mathrm{A}}{\mathrm{z}}\right)=\dfrac{220}{84}=2.62$

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

$$ \begin{aligned} & A _{1}=(220)-(4 \times 3)=208\\ & z _{1}=[(84)-(4 \times 3)]+(4 \times 1)=82 \end{aligned} $$

$\therefore$ Final value of $\dfrac{\mathrm{A} _{1}}{\mathrm{z} _{1}}=\dfrac{208}{82} \simeq 2.54$

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

Hence option(1) is correct.

Correct answer: (3)

Solution:

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

When one atom of $X$ disintegrates, one atom of $Y$ is produced. $A t t=t$; the rate of production $Y$, equals the rate of disintegration of $\mathrm{X}\left(=\lambda _{1} \mathrm{~N} _{1}\right)$. The instantaneous rate of decrease of $\mathrm{Y}$, due to its own decay into $\mathrm{Z}$, is $\left(-\lambda _{2} N _{2}\right)$. The net rate of change of the number of along of $Y$, is

$\mathrm{R} _{\mathrm{Y}}=\dfrac{\mathrm{dN} _{2}}{\mathrm{dt}}=\lambda _{1} \mathrm{~N} _{1}-\lambda _{2} \mathrm{~N} _{2}$ $\hspace{5 cm}$ …….(2)

$\therefore \dfrac{\mathrm{R} _{\mathrm{X}}}{\mathrm{R} _{\mathrm{Y}}}=\dfrac{-\lambda _{1} \mathrm{~N} _{1}}{\lambda _{1} \mathrm{~N} _{1}-\lambda _{2} \mathrm{~N} _{2}}=\dfrac{1}{\left(\dfrac{\lambda _{2} \mathrm{~N} _{2}}{\lambda _{1} \mathrm{~N} _{1}}-1\right)}=\left(\dfrac{\lambda _{2} \mathrm{~N} _{2}}{\lambda _{1} \mathrm{~N} _{1}}-1\right)^{-1}$

Hence option(3) is correct.

Correct answer: (3)

Solution:

The nuclear reaction is $4\left({ } _{2}^{4} \mathrm{He}\right) \rightarrow{ } _{8}^{16} \mathrm{O}$

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

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

Hence option (3) is correct.

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) ${ } _{3}^{7} \mathrm{Li} \rightarrow{ } _{2}^{4} \mathrm{He}+{ } _{1}^{3} \mathrm{H}$

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 $\left(m _{3}+m _{4}\right)>M _{1}$; the reaction is not possible

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

$\text { Mass of }{ } _{82}^{206} \mathrm{~Pb}=\mathrm{M} _{1}=205.974454 \mathrm{u}$

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) ${ } _{1}^{2} \mathrm{H}+{ } _{2}^{4} \mathrm{He} \rightarrow{ } _{3}^{6} \mathrm{Li}$

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

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 $\left(m _{1}+m _{2}\right)<M$, this reaction is not possible

Hence option (3) is correct.

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.

Correct answer: (3)

Solution:

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

The total mass energy of electron position pair

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

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

$$ \begin{aligned} & \mathrm{h} \nu _{\mathrm{m}}=163.8 \times 10^{-15} \mathrm{~J}\\ \therefore \quad & \nu _{\mathrm{m}}=\dfrac{163.8 \times 10^{-15}}{6.6 \times 10^{-34}} \mathrm{~Hz} \simeq 2.482 \times 10^{20} \mathrm{~Hz} \simeq 2.5 \times 10^{20} \mathrm{~Hz} \end{aligned} $$

Hence option(3) is correct.

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 ${ } _{92}^{235} \mathrm{U}=\dfrac{6.023 \times 10^{23}}{0.256 \times 235} \simeq 10^{22}$

$\therefore$ 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, $\eta=\dfrac{E _{\text {out }}}{E _{\text {in }}}=0.8$

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

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

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

Hence option (2) is correct.

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

$$ \begin{aligned} & \dfrac{1}{2} \mathrm{mv}^{2}=\dfrac{\mathrm{q} _{\alpha}(\mathrm{Ze})}{4 \pi \varepsilon _{0} \mathrm{r} _{0}}\\ & \mathrm{r} _{0}=\dfrac{2 \mathrm{q} _{\alpha}(\mathrm{Ze})}{4 \pi \varepsilon _{0} \mathrm{mv}^{2}} \end{aligned} $$

Hence $\mathrm{r} _{0} \propto \dfrac{1}{\mathrm{~m}}$

Hence option (3) is correct.

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 $\dfrac{217.6}{13.6}=\dfrac{\mathrm{z}^{2}}{1}$

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

Hence $\mathrm{z}=4$

Hence option (1) is correct

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 $\mathrm{T}=\dfrac{2 \pi \mathrm{r}}{\mathrm{v}}$

Now from Bohr’s theory we have

$$ \mathrm{r} \propto \mathrm{n}^{2} \text { and } \mathrm{v} \propto \dfrac{1}{\mathrm{n}} $$

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

$$ \left(\dfrac{\mathrm{T} _{1}}{\mathrm{~T} _{2}}\right)^{2}=\left(\dfrac{\mathrm{R}}{4 \mathrm{R}}\right)^{3}=\dfrac{1}{64} \text { or } \dfrac{\mathrm{T} _{1}}{\mathrm{~T} _{2}}=\dfrac{1}{8} $$

Hence option (2) is correct

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.

$$ \mathrm{E}=-13.6 \dfrac{\mathrm{z}^{2}}{\mathrm{n}^{2}} $$

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

$$ \mathrm{E}=\dfrac{-13.6 \times(3)^{2}}{(1)^{2}} \mathrm{eV}=122.4 \mathrm{eV} $$

Hence option (4) is correct.

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

$$ \begin{aligned} & \dfrac{1}{\lambda _{\mathrm{H} _{1}}}=\mathrm{R} \times 1^{2}\left(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right)=\mathrm{R}\left(\dfrac{1}{4}-\dfrac{1}{9}\right)\\ & \lambda _{\mathrm{H} _{1}}=\dfrac{36}{5 \mathrm{R}} \end{aligned} $$

or $\quad 6561=\dfrac{36}{5 \mathrm{R}}$

The second line of Balmer series of ionzied helium atom corresponds to transition from $n _{1}=4$ to $n _{2}=2$

Hence $\dfrac{1}{\lambda _{\text {He }}(2)}=\mathrm{R} \times 2^{2}\left(\dfrac{1}{2^{2}}-\dfrac{1}{4^{2}}\right)$

$$ \begin{aligned} & =4 \mathrm{R}\left(\dfrac{1}{4}-\dfrac{1}{6}\right)=4 \mathrm{R}\left(\dfrac{3}{16}\right)=\dfrac{3 \mathrm{R}}{4}\\ \lambda _{\mathrm{He} _{(2)}} & =\dfrac{4}{3 \mathrm{R}}\\ \dfrac{\lambda _{\mathrm{He} _{(2)}}}{\lambda _{\mathrm{H} _{1}}} & =\dfrac{4}{3 \mathrm{R}} \times \dfrac{5 \mathrm{R}}{36}=\dfrac{5}{27}\\ \lambda _{\mathrm{He} _{(2)}} & =\dfrac{6561 \times 5}{27}=1215 \AA \end{aligned} $$

Hence option(1) is correct.

Correct answer: (3)

Solution:

The energy expression corresponding to $\mathrm{n}^{\text {th }}$ permitted level is given by $\mathrm{E} _{\mathrm{n}}=-\dfrac{\mathrm{Rhc}}{\mathrm{n}^{2}}$

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 $E _{n}=-\dfrac{R h c}{n^{2}}$

Energy of the hypothetical particle will be $\mathrm{E} _{\mathrm{n}}=-\dfrac{2 \mathrm{Rhc}}{\mathrm{n}^{2}}$

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 $n=2$.

$$ \therefore \quad \dfrac{\mathrm{hc}}{\lambda _{\max }}=2 \operatorname{Rhc}\left(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right) $$

which gives $\lambda _{\max }=\dfrac{18}{5 \mathrm{R}}=3.6 \mathrm{R}$

Hence option(3) is correct.

Correct answer: (2)

Solution:

We known that the ground state energy of the hydrogen atom is gien by $E _{1}=-\dfrac{2 \pi^{2} \mathrm{~K}^{2} \mathrm{~m} _{\mathrm{e}} \mathrm{e}^{4}}{\mathrm{~h}^{2}}$

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

$$ \begin{aligned} & \mathrm{E}(\text { mounic })=\mathrm{E} _{1}=-\dfrac{2 \pi^{2} \mathrm{~K}^{2} \mathrm{~m} _{\mathrm{e}} \mathrm{e}^{4}}{\mathrm{~h}^{2}} \times 207=-13.6 \times 207 \mathrm{eV}\\ & =-2.8 \times 10^{3} \mathrm{eV} \end{aligned} $$

Hence option (2) is correct.

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 $V=\dfrac{E _{2}-E _{1}}{e}=\dfrac{h c}{e} R\left(\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right)$

$$ \mathrm{V}=\dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19}}=10.22 \mathrm{~V} $$

Hence option(2) is correct.

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 $4 \rightarrow 3 \rightarrow \mathrm{E} _{2}-\mathrm{E} _{1}=\mathrm{R}\left(\dfrac{1}{2^{2}}-\dfrac{1}{4^{2}}\right)$ ultraviolet

$$ =\mathrm{R}\left(\dfrac{1}{9}-\dfrac{1}{16}\right)=\dfrac{7}{128} \mathrm{R}=(0.0546) \mathrm{R} $$

For transition $2 \rightarrow 1 \rightarrow\left(\mathrm{E} _{2}-\mathrm{E} _{1}\right)=\mathrm{R}\left(1-\dfrac{1}{4}\right)=\dfrac{3}{4} \mathrm{R}=0.75 \mathrm{R}$

For transition $3 \rightarrow 2 \rightarrow\left(\mathrm{E} _{2}-\mathrm{E} _{1}\right)=\mathrm{R}\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=\dfrac{5}{36} \mathrm{R}=0.088 \mathrm{R}$

For transition $4 \rightarrow 2 \rightarrow \mathrm{R}\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{16} \mathrm{R}=0.1875 \mathrm{R}$

For transition $5-4 \rightarrow R\left(\dfrac{1}{16}-\dfrac{1}{25}\right)=\dfrac{9}{400} R=0.225 R$

From this data we find that option (4) is the correct option.

Correct answer: (1)

Solution:

According to Bohr’s theory we have $\bar{v}=\dfrac{1}{\lambda}=R\left(\dfrac{1}{1^{2}}-\dfrac{1}{n^{2}}\right)$

or $\dfrac{\mathrm{R}}{\mathrm{n}^{2}}=\left(\mathrm{R}-\dfrac{1}{\lambda}\right)=\dfrac{\lambda \mathrm{R}-1}{\lambda}$

$\begin{aligned} & \mathrm{n}^{2}=\dfrac{\lambda \mathrm{R}}{(\lambda \mathrm{R}-1)}\\ & \mathrm{n}=\sqrt{\dfrac{\lambda \mathrm{R}}{(\lambda \mathrm{R}-1)}} \end{aligned}$

Hence option(1) is correct.

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 $\dfrac{n(n-1)}{2}$

$\therefore$ The number of lines that would be emitted when a transition take place from $n=4$ to ground state is given by $\dfrac{4(4-1)}{2}=6$

Hence option(4) is correct.

Correct answer: (4)

Solution:

According to Bohr’s theory of hydrogen atom, frequency of radiation emitted is given by relation

$$ \mathrm{h} \nu=\mathrm{R}\left[\dfrac{1}{(\mathrm{n}-1)^{2}}-\dfrac{1}{\mathrm{n}^{2}}\right] $$

where $\mathrm{R}$ is Rydberg like constant for hydrogen like atom

$$ \begin{aligned} \nu & =\dfrac{1}{h} R\left[\dfrac{n^{2}-(n-1)^{2}}{n^{2}(n-1)^{2}}\right]\\ \text { or } \quad \nu & =\dfrac{R}{h}\left[\dfrac{(n+n-1)(n+n+1)}{n^{2}(n-1)^{2}}\right]\\ \nu & =\dfrac{R}{h}\left[\dfrac{2 n-1}{n^{2}(n-1)^{2}}\right] \end{aligned} $$

Since $\mathrm{n}»1$ we have $\nu \simeq \dfrac{\mathrm{R}}{\mathrm{h}}\left(\dfrac{2 \mathrm{n}}{\mathrm{n}^{4}}\right)=\dfrac{\mathrm{R}}{\mathrm{h}}\left(\dfrac{1}{\mathrm{n}^{3}}\right)$

or $\quad \nu \propto \dfrac{1}{\mathrm{n}^{3}}$

Hence option(4) is correct.

Correct answer: (4)

Solution:

For transition $\mathrm{n} _{2}=3$ to $\mathrm{n} _{1}=2$ we have

$$ \dfrac{\mathrm{hc}}{\lambda _{1}}=\mathrm{E} _{3}-\mathrm{E} _{4}=13.6\left[\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right]=13.6 \times \dfrac{1}{44} $$

For transitionto $\mathrm{n} _{2}=2$ to $\mathrm{n} _{1}=1$ we similarly have

$$ \dfrac{\mathrm{hc}}{\lambda _{2}}=\mathrm{E} _{2}-\mathrm{E} _{3}=13.6\left[\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right] $$

From above two equations we have

$$ \dfrac{\lambda _{1}}{\lambda _{2}}=\dfrac{5 / 36}{7 / 144}=\dfrac{20}{7} $$

Hence option (4) is correct.

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.

Correct answer: (2)

Solution:

For Lyman series, the wavelength of lines is obtained from the relation $\bar{v}=\dfrac{1}{\lambda}=\mathrm{R}\left(\dfrac{1}{1^{2}}-\dfrac{1}{\mathrm{n}^{2}}\right)$

We will get a line with shortest wavelength corresponding to transition from $n=\infty$ to $n=1$

Hence $\dfrac{1}{\lambda _{\mathrm{s}}}=\mathrm{R}\left(\dfrac{1}{1}-\dfrac{1}{\mathrm{n}^{2}}\right)=\mathrm{R}$

$$ \lambda _{\mathrm{S}}=\dfrac{1}{\mathrm{R}}=\dfrac{10^{10}}{10967700}=911 \AA $$

For longest wavelength we will have $\mathrm{n}=2$

Hence $\lambda _{\mathrm{L}}=\mathrm{R}\left(1-\dfrac{1}{4}\right)=\dfrac{3}{4} \mathrm{R}$

$$ \lambda=\dfrac{4}{3} \mathrm{R}=\dfrac{4 \times 10^{10}}{3 \times 10967700}=1215 \AA $$

Hence option (2) is correct.

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

$$ \bar{v} _{\alpha}=\dfrac{1}{\lambda _{\alpha}}=\mathrm{R}\left(\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right)=\mathrm{R}\left(\dfrac{1}{9}-\dfrac{1}{16}\right) $$

or $\quad \lambda _{\alpha}=\dfrac{144}{7 \mathrm{R}}$

Similarly $\mathrm{H} _{\beta}$ line will be obtained corresponding to $\mathrm{n} _{1}=3$ and $\mathrm{n} _{2}=5$

$$ \begin{aligned} & \bar{v} _{\beta}=\dfrac{1}{\lambda _{\beta}}=\mathrm{R}\left(\dfrac{1}{3^{2}}-\dfrac{1}{5^{2}}\right)=\mathrm{R}\left(\dfrac{1}{9}-\dfrac{1}{25}\right)=\dfrac{16}{225} \mathrm{R}\\ & \lambda _{\beta}=\dfrac{225}{16 \mathrm{R}}\\ & \dfrac{\lambda _{\alpha}}{\lambda _{\beta}}=\dfrac{144}{7 \mathrm{R}} \times \dfrac{16 \mathrm{R}}{225} \approx 1.5 \end{aligned} $$

Hence option (3) is correct.

Correct answer: (4)

Solution:

Since the series involved is Balmer series, the transition terminates at $\mathrm{n}=2$

$$ \therefore \quad \bar{v}=\dfrac{1}{\lambda}\left(\dfrac{1}{\mathrm{n} _{1}^{2}}-\dfrac{1}{\mathrm{n} _{2}^{2}}\right) $$

It is clear that $\lambda$ will have maximum value corresponding to transition from $n _{2}=3$ to $n _{1}=2$

$$ \therefore \dfrac{1}{\lambda _{\max }}=\mathrm{R}\left(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right)=\mathrm{R}\left(\dfrac{5}{36}\right) $$

For minimum value of $\lambda$ the transition has to be from $\infty$ to $n _{1}=2$

$$ \begin{aligned} & \dfrac{1}{\lambda _{\min }}=\mathrm{R}\left(\dfrac{1}{2^{2}}-\dfrac{1}{\infty}\right)=\mathrm{R} \dfrac{1}{4}\\ & \dfrac{\lambda _{\min }}{\lambda _{\max }}=\dfrac{4}{\mathrm{R}} \times \dfrac{5 \mathrm{R}}{36}=\dfrac{5}{9} \end{aligned} $$

Hence option(4) is correct.

Correct answer: (2)

Solution:

According to the relation $\bar{v}=\dfrac{1}{\lambda}=\mathrm{R}\left(\dfrac{1}{\mathrm{n} _{1}^{2}}-\dfrac{1}{\mathrm{n} _{2}^{2}}\right)$ 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

$$ \dfrac{1}{\lambda _{\mathrm{L}}}=\mathrm{R}\left(\dfrac{1}{1^{2}}-\dfrac{1}{2^{2}}\right)=\mathrm{R}\left(\dfrac{3}{4}\right) $$

and for Balmer series

$$ \begin{aligned} & \dfrac{1}{\lambda _{\mathrm{B}}}=\mathrm{R}\left(\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right)=\mathrm{R}\left(\dfrac{5}{36}\right)\\ & \dfrac{\lambda _{\mathrm{L}}}{\lambda _{\mathrm{B}}}=\dfrac{5 \mathrm{R}}{36} \times \dfrac{4}{3 \mathrm{R}}=\dfrac{5}{27} \end{aligned} $$

Hence option(2) is correct.

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

$$ \bar{v}=\dfrac{1}{\lambda} \mathrm{R}\left(\dfrac{1}{\mathrm{n} _{1}^{2}}-\dfrac{1}{\mathrm{n} _{2}^{2}}\right) $$

$$ \dfrac{1}{\lambda}=\mathrm{R}\left(\dfrac{1}{1^{2}}-\dfrac{1}{5^{2}}\right)=\dfrac{24 \mathrm{R}}{25} \text { or } \lambda=\dfrac{25}{24 \mathrm{R}} $$

Linear momentum of photon emitted is $p=\dfrac{h}{\lambda}=\dfrac{h \times 24 R}{25}$

Now $\mathrm{p}=\mathrm{m} \nu \quad(\because$ hydrogen atom is stationary $)$

$$ \mathrm{m} \nu=\dfrac{24 \mathrm{Rh}}{25} \quad \nu=\dfrac{24 \mathrm{hR}}{25 \mathrm{M}} $$

Hence option (2) is correct.

Correct answer: (4)

Solution:

Bohr’s condition for quantization of angular momentum is $\mathrm{mvr}=\dfrac{\mathrm{nh}}{2 \pi}$

where $\mathrm{m}$ is the reduced mass of diatomic molecule and is $=\dfrac{\mathrm{m} _{1}+\mathrm{m} _{2}}{\mathrm{~m} _{1} \mathrm{~m} _{2}}$

If $\mathrm{v}$ denotes linear velocity and $\omega$ the angular velocity we have $\mathrm{v}=\mathrm{r} \omega$

$\therefore \mathrm{m}(\mathrm{r} \omega) \mathrm{r}=\mathrm{n} \hbar$ or $\omega=\dfrac{\mathrm{n} \hbar}{\mathrm{mr}^{2}}$

The rotational energy of the given diatomic molecule is given by

$$ \dfrac{1}{2} \mathrm{I} \omega^{2}=\dfrac{1}{2}\left(\mathrm{mr}^{2}\right)\left(\dfrac{\mathrm{n} \hbar}{\mathrm{mr}^{2}}\right)^{2} $$

$\therefore$ Rotational energy is $\dfrac{\mathrm{n}^{2} \hbar^{2}}{2 \mathrm{mr}^{2}}=\dfrac{\mathrm{n}^{2} \hbar^{2}\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right)}{2 \mathrm{~m} _{1} \mathrm{~m} _{2} \mathrm{r}^{2}}$

Hence option (4) is correct.

Correct answer: (3)

Solution:

When a transition take place from state $\mathrm{n}=3$ to $\mathrm{n}=1$, the energy of photon emitted is

$$ \mathrm{E}=\mathrm{E} _{2}-\mathrm{E} _{1}=\dfrac{-13.6}{3^{2}}-\left(\dfrac{-13.6}{1^{2}}\right)=12.1 \mathrm{eV} \text { which equals to h$v$ } $$

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.

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 $-\dfrac{13.6 \times 9}{(3)^{2}} \mathrm{eV}=-13.6 \mathrm{eV}$

$\therefore \quad$ 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

$$ \lambda=\dfrac{\mathrm{hc}}{\mathrm{E} _{3}-\mathrm{E} _{1}}=\dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{13.6 \times 8 \times 1.6 \times 10^{-19}}=0.11426 \times 10^{-7} \mathrm{~m}=114.26 \mathrm{~A} $$

Hence option(2) is correct.

Correct answer: (1)

Solution:

We know the energy of an electron in th $n^{\text {th }}$ orbit of hydrogen atom is given by $E _{n}=-\dfrac{13.6}{n^{2}} \mathrm{eV}$ for $\mathrm{n}=1$ (the ground state) $\mathrm{E} _{1}=-13.6 \mathrm{eV}$

for $\mathrm{n}=4 \mathrm{E} _{4}=-\dfrac{13.6}{16}=-0.85 \mathrm{eV}$

Energy of photon $=\mathrm{E} _{4}-\mathrm{E} _{1}=-0.85-(-13.6)=12.75 \mathrm{eV}$ and this must be equal to $\dfrac{\mathrm{hc}}{\lambda}$

$$ \begin{aligned} & \dfrac{\mathrm{hc}}{\lambda}=12.75 \mathrm{eV}\\ & \lambda=\dfrac{\mathrm{hc}}{12.75}=\dfrac{4.14 \times 10^{-15} \times 3 \times 10^{8}}{12.75}=974 \mathrm{~A} \end{aligned} $$

Hence option(1) is correct.

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

$$ 2.7<\mathrm{z}^{2}=\left(\mathrm{E} _{2}-\mathrm{E} _{1}\right)<\mathrm{z}^{2}=\left(\mathrm{E} _{3}-\mathrm{E} _{2}\right) $$

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.

Correct answer: (1)

Solution:

Let $\lambda$ be required wavlength

Then we have $\mathrm{eV}=\dfrac{\mathrm{hc}}{\lambda}$ where $\mathrm{V}$ denotes the corresponding critical potential

$\therefore \quad \lambda=\dfrac{\mathrm{hc}}{\mathrm{eV}}=\dfrac{6.61 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 13.05}=9515 \mathrm{~A}$

Hence option (4) is correct.

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.

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.

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 $\dfrac{3}{4}$.

Hence option(2) is correct.

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

$$ \lambda _{1}=\dfrac{0.693}{\mathrm{~T} _{1}} \text { and } \lambda _{2}=\dfrac{0.693}{\mathrm{~T} _{2}} $$

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 $\left(\lambda _{1} \mathrm{dt}+\lambda _{2} \mathrm{dt}\right)$. 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 $\dfrac{0.693}{\mathrm{~T}}=\dfrac{0.693}{\mathrm{~T} _{1}}+\dfrac{0.693}{\mathrm{~T} _{2}}$

or $\dfrac{1}{\mathrm{~T}}=\dfrac{1}{\mathrm{~T} _{1}}+\dfrac{1}{\mathrm{~T} _{2}}$

Hence option (2) is correct.

Correct answer: (4)

Solution:

$\because \quad \mathrm{N}=\mathrm{N} _{0} \mathrm{e}^{-\lambda \mathrm{t}}$ hence $\left|\dfrac{\mathrm{dN}}{\mathrm{dt}}\right|=\left|\lambda \mathrm{N} _{0} \mathrm{e}^{-\lambda \mathrm{t}}\right|$

$$ \begin{aligned} & \ell n\left(\dfrac{d N}{d t}\right)=\ell n _{e}\left(\lambda N _{0}\right)-\lambda t\\ & \ell n\left|\dfrac{d N}{d t}\right|=-\lambda t+\log _{e}\left(\lambda N _{0}\right) \end{aligned} $$

which is of the form of a straight line $y=m x+c$ whose slope is $-\lambda=\dfrac{3-4}{6-4}$

or $\quad \lambda=\left(\dfrac{1}{2}\right)$ year $^{-1}$

Half life $\mathrm{T}=\dfrac{0.693}{\lambda}=2 \times 0.693=1.386$ years

Now $\mathrm{N}=\mathrm{N} _{0}\left(\dfrac{1}{2}\right)^{1 / 2}$ or $\quad \dfrac{\mathrm{N}}{\mathrm{N} _{0}}=\left(\dfrac{1}{2}\right)^{1 / T}$

$$ \begin{aligned} \dfrac{1}{p} & =\left(\dfrac{1}{2}\right) \dfrac{4.16}{1.386}=\left(\dfrac{1}{2}\right)^{3}=\dfrac{1}{8}\\ \text { or } \quad p & =8 \end{aligned} $$

Hence option (4) is correct.

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 $=\dfrac{10 \mathrm{mg}}{(2)^{2}}=2.5 \mathrm{mg}$

Hence option (2) is correct.

Correct answer: (2)

Solution:

Initial counting rate $=47.5 \mathrm{a}$ particles $/ \mathrm{min}$

Final counting rate after 5 minutes $27 \alpha$ particles/min

$\therefore \quad 27=47.5 \mathrm{e}^{-\lambda}$ or $\dfrac{47.5}{27}=\mathrm{e}^{\lambda}$

or $\quad \lambda=\dfrac{2.303 \times 0.2455}{5}=0.1157 \mathrm{~min}^{-1}$

Half life of sample $\mathrm{T} _{1 / 2}=\dfrac{0.693}{\lambda}=\dfrac{0.693}{0.1157} \mathrm{~min}=5.989 \mathrm{~min} \simeq 5.99$ minute

Hence option (2) is correct.

Correct answer: (4)

Solution:

Number of $\alpha$ particle per sec in all direction is given by $\dfrac{22 \times 4 \pi}{60}=\dfrac{22 \pi}{15}$

No. of atoms in 1 milligram of thorium $=\dfrac{0.001 \times 6.02 \times 10^{23}}{232}$

Decay rate per second $=\dfrac{22 \pi}{15}$

$$ \begin{aligned} & \therefore \quad-\dfrac{22 \pi}{15}=-\dfrac{\lambda \times 0.001 \times 6.02 \times 10^{23}}{232}\\ & \quad \lambda=\dfrac{232 \times 22 \pi}{15 \times 0.001 \times 6.023 \times 10^{23}}=1.775 \times 10^{-12}\\ & \quad \mathrm{~T} _{1 / 2}=\dfrac{0.693}{\lambda}-1.23 \times 10^{10} \text { year } \end{aligned} $$

Hence option (4) is correct.

Correct answer: (1)

Solution:

Dose per $\mathrm{ml}=\dfrac{5}{3500} \mathrm{mc}$

Number of disintegration $\dfrac{1}{700} \times 3.7 \times 10^{7}$ per sec. in blood

After one hour we can assume no decay.

Disintegration counted $=\dfrac{1}{10} \times \dfrac{1}{700} \times 3.7 \times 10^{7}=5.3 \times 10^{2} \mathrm{dis} / \mathrm{s}$

Hence option(1) is correct.

Correct answer: (4)

Solution:

In $\beta$ decay process atomic number increases by 1 whereas mass number remains same. Hence the possible nuclear reaction is

$$ { } _{29} \mathrm{Cu}^{64} \rightarrow{ } _{30} \mathrm{Zn}^{64}+{ } _{-1} \mathrm{e}^{0} $$

Hence option (4) is correct.

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.

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

$$ \dfrac{\mathrm{N}}{\mathrm{N} _{0}}=\dfrac{1}{32}=\dfrac{1}{(2)^{5}} $$

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.

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

$$ \lambda=\dfrac{0.693}{5700} \text { year }^{-1} \hspace{4 cm} \left(\because \mathrm{T} _{1 / 2}=5700 \text { years }\right) $$

Representing by $t$ the age of ancient wood we have

$$ \begin{aligned} & 50=200 e^{-\left(\dfrac{0.693}{5700}\right) \mathrm{t}} \text { or } 4=\mathrm{e}^{\dfrac{0.693}{5700} \mathrm{t}}\\ & \mathrm{t}=\dfrac{2.302 \times 6021 \times 5700}{0.693}=1.41 \times 10^{4} \text { years } \end{aligned} $$

Hence option (3) is coorect.

Correct answer: (2)

Solution:

The following reaction represents $\beta$ decay process

$$ \mathrm{n} \rightarrow \mathrm{p}+\overline{\mathrm{e}}+\overline{\mathrm{v}} $$

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.

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

$$ \mathrm{E}=\mathrm{h} v=\mathrm{E} _{\mathrm{n} _{2}}-\mathrm{E} _{\mathrm{n} _{1}}=\mathrm{R}\left(\dfrac{1}{\mathrm{n} _{1}^{2}}-\dfrac{1}{\mathrm{n} _{2}^{2}}\right) $$

Also the value of $E$ corresponding to quantum number $n$ is given by $E _{n}=-\dfrac{13.6}{n^{2}} e V$

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.

Correct answer: (3)

Solution:

During a negative beta decay, the reaction which takes place is

$$ { } _{0}^{1} \mathrm{n} \rightarrow{ } _{1}^{1} \mathrm{p}+{ } _{-1}^{0} \mathrm{e}+\bar{v} $$

Here $\bar{v}$ is an anti neutrinos

Hence option(3) is correct.

Correct answer: (2)

Solution:

The mass of neutron is given by

$$ \mathrm{M}\left({ } _{1} \mathrm{H}^{2}\right)-\mathrm{M}\left({ } _{1} \mathrm{H}^{1}\right)-\dfrac{2.27}{931} \mathrm{amu} $$

$$ =-0015380+1.008145-0.002392=1.0089999 \mathrm{~amu} $$

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.

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.

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.

Correct answer: (2)

Solution:

In a nuclear reaction, the mass number and atomic number of reactants and products are conserved.

Hence

$$ \begin{aligned} & \mathrm{Ne} _{10}^{22}+\text { energy } \rightarrow 2 \mathrm{He} _{2}^{4}+\mathrm{X}\\ & \mathrm{Ne} _{10}^{22} \rightarrow 2 \mathrm{He} _{2}^{4}+\mathrm{X} _{6}^{14} \end{aligned} $$

Obvisouly $\mathrm{X}$ is carbon with mass number 14 and atomic number 6 .

Hence option (2) is correct.

Correct answer: (4)

Solution:

The reaction

$$ \mathrm{P}^{238} \rightarrow \mathrm{U}^{234}+{ } _{2} \mathrm{He}^{4}+\mathrm{Q} $$

Represents the decay process of $\mathrm{Pu}^{238}$

Here $Q$ is representing the energy of nuclear reaction

$\mathrm{Q}=\mathrm{m}\left(\mathrm{Pu}^{238}\right)-\left[\mathrm{m}\left(\mathrm{U}^{234}\right)+\mathrm{m}\left({ } _{2} \mathrm{He}^{4}\right)\right]$

$\mathrm{Q}=238.04954-[234.04096+4.002603]=0.00598 \mathrm{amu}=0.00598 \times 931=5.57 \mathrm{MeV}$

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.

Correct answer: (1)

Solution:

Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei. Hence option (1) is correct.

Correct answer: (4)

Solution:

The reaction is represented by $2{ } _{1} \mathrm{H}^{2} \rightarrow{ } _{2} \mathrm{He}^{4}+$ energy

Mass defect $\Delta \mathrm{m}[2 \times 2.04178-4.00388]=0.02568$ amuu

Energy released $=0.02568 \times 931 \mathrm{MeV}=23.91 \mathrm{MeV}$

Since efficiency of the process is $30 %$ actual output is $23.91 \times \dfrac{30}{100}=7.173 \mathrm{MeV}$

Actual output per deutron atom is $\dfrac{7.173}{2}=3.587 \mathrm{MeV}=3.587 \times 1.6 \times 10^{-12} \mathrm{~J}$

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

$$ =\dfrac{5 \times 10^{7}}{3.587 \times 1.6 \times 10^{-12}}=871.4 \times 10^{17} \approx 10^{20} \mathrm{atoms} / \mathrm{s} $$

Hence option (4) is correct.

Correct answer: (4)

Solution:

Total mass defect $\Delta \mathrm{m}$ 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.

$\therefore \quad$ Energy realeased per nucleus $=\dfrac{0.02866 \times 931}{4}=6.67 \mathrm{MeV}$

Hence option (4) is correct.

Correct answer: (3)

Solution:

The given reactions are

$$ \begin{aligned} & { } _{1}^{2} \mathrm{H}+{ } _{1}^{2} \mathrm{H} \rightarrow{ } _{1}^{3} \mathrm{H}+\mathrm{p}\\ & { } _{1}^{2} \mathrm{H}+{ } _{1}^{3} \mathrm{H} \rightarrow{ } _{2}^{4} \mathrm{He}+\mathrm{n}\\ & 3\left({ } _{1}^{2} \mathrm{H}\right)+{ } _{2}^{4} \mathrm{He}+\mathrm{n}+\mathrm{p} \end{aligned} $$

Mass defect $\Delta \mathrm{m}=(3 \times 2.014-4.001-1.007-1.0008)=0.026 \mathrm{amu}$

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 $=\dfrac{10^{40} \times 3.87 \times 10^{-12}}{3}=1.29 \times 10^{28} \mathrm{~J}$

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 $\dfrac{1.29 \times 10^{28}}{10^{16}}=1.29 \times 10^{12} \mathrm{~s} \approx 10^{12}$

Hence option (3) is correct.