exemplar-phy-12-atoms

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Ques. The binding energy of a $H$-atom, considering an electron moving around a fixed nuclei (proton), is $B=-\frac{m e^{4}}{8 n^{2} \varepsilon_0^{2} h^{2}}$ (M = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be
$B=-\frac{m e^{4}}{8 n^{2} \varepsilon_0^{2} h^{2}}$ (M = proton mass)
This last expression is not correct, because
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(a) $n$ would not be integral 
(b) Bohr-quantisation applies only two electron 
(c) the frame in which the electron is at rest is not inertial 
(d) the motion of the proton would not be in circular orbits, even approximately.
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Ques. For the ground state, the electron in the $H$-atom has an angular momentum $=h$, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,
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(a) because Bohr model gives incorrect values of angular momentum
(b) because only one of these would have a minimum energy
(c) angular momentum must be in the direction of spin of electron
(d) because electrons go around only in horizontal orbits
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Thinking Process
Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 \pi}$ where $h$ is the Planck’s constant $(=6.6 \times 10^{-34} Js)$.
Ans. (a)
In the simple Bohr model, only the magnitude of angular momentum is kept equal to some integral multiple of $\frac{h}{2 \pi}$, where, $h$ is Planck’s constant and thus, the Bohr model gives incorrect values of angular momentum.
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Thinking Process
The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus.
This happens only because the nuclear force is much stronger than the Coulomb force. The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres.
Ans. (a)
In the molecules, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged.
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Thinking Process
The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0. The energy of this state $(n=1), E_1$ is -13.6 eV.
Ans. (a)
The total energy associated with the two $H$-atoms in the ground state collide in elastically $=2 \times(13.6 eV)=27.2 eV$.
The maximum amount by which their combined kinetic energy is reduced when any one of them goes into first excited state after the inelastic collision.
<Success>Sol. to be continued …</Success>
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Ques. The Balmer series for the $H$-atom can be observed
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(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state
(c) in any transition in a $H$-atom
(d) as a sequence of frequencies with the higher frequencies getting closely packed
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Thinking Process
The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines.
Ans. (b, d)
Balmer series for the $H$-atom can be observed if we measure the frequencies of light emitted due to transitions between higher excited states and the first excited state and as a sequence of frequencies with the higher frequencies getting closely packed.
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Ques. Let $E_n=\frac{-1 m e^{4}}{8 \varepsilon_0^{2} n^{2} h^{2}}$ be the energy of the $n$th level of $H$-atom. If all the $H$-atoms are in the ground state and radiation of frequency $\frac{(E_2-E_1)}{h}$ falls on it,
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(a) it will not be absorbed at all
(b) some of atoms will move to the first excited state
(c) all atoms will be excited to the state
(d) no atoms will make a transition to the $n=3$ state
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Thinking Process
When an atom absorbs a photon that has precisely the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption.
Ans. (b, d)
When all the $H$-atoms are in the ground state and radiation of frequency $\frac{(E_2-E_1)}{h}$ falls on it, some of atoms will move to the first excited state and no atoms will make a transition to the $n=3$ state.
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Thinking Process
Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy, say kinetic energy and vice-versa. Einstein gave the famous mass-energy equivalence relation $E=m c^{2}$ where the energy equivalent of mass $m$ is related by the above equation and $c$ is the velocity of light.
Ans.
Since, the difference in mass of a nucleus and its constituents, $\Delta M$, is called the mass defect and is given by
$\Delta M=[Z m_p+(A-Z) m_n]-M$
<Success>Sol. to be continued …</Success>
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Thinking Process
Neil’s Bohr proposed a model for hydrogenic (single electron) atoms in order to explain the stability of atoms.
Ans.
On removing one electron from ${} _{2}He^{4}$ and ${} _{2}He^{3}$, the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass.Also after removing one electron from ${} _{2}He^{4}$ and ${} _{2}He^{3}$ atoms contain one electron and are hydrogen like atoms.
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Thinking Process
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons provides necessary centripetal force of revolution. Also, the magnitude of electrostatic force $F \propto q_1 q_2$.
Ans.
If proton had a charge $(\frac{+4}{3}) e$ and electron a charge $(\frac{-3}{4}) e$, then the Bohr formula for the $H$-atom remain same, since the Bohr formula involves only the product of the charges which remain constant for given values of charges.
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Thinking Process
Bohr’s postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of $\frac{h}{2 \pi}$, where $h$ is Planck’s constant $(=6.6 \cdot 10^{-34} J-s)$. Thus, the angular momentum $(L)$ of the orbiting electron is quantised. i.e.,
$L=\frac{n h}{2 \pi}$
Ans.
According to Bohr model electrons having different energies belong to different levels having different values of $n$. So, their angular momenta will be different, as
$L=\frac{n h}{2 \pi} \text{ or } L \propto n$
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Thinking Process
The total energy of the electron in the nth stationary states of the hydrogen. Atom of hydrogen like atom of atomic number $Z$ is given by
$
E_n=z^{2} \frac{-13.6 eV}{n^{2}}
$
Ans.
For a $He$-nucleus with charge $2 e$ and electrons of charge $-e$, the energy level in ground state is
$-E_n=Z^{2} \frac{-13.6 eV}{n^{2}}=2^{2} \frac{-13.6 eV}{1^{2}}=-54.4 eV$
Thus, the ground state will have two electrons each of energy $E$ and the total ground state energy would be $-(4 \times 13.6) eV=-54.4 eV$.
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Thinking Process
The electric current due to revolution of charge is given by $i=\frac{9}{T}=Q \frac{1}{T}=Q \times n$, where $n$ is frequency.
Ans.
The electron in Hydrogen atom in ground state revolves on a circular path whose radius is equal to the Bohr radius $(a_n)$. Let the velocity of electron is $v$.
$\therefore$ Number of revolutions per unit time $=\frac{2 \pi a_0}{v}$
The electric current is given by $i=\frac{q}{t}$, if $q$ charge flows in time $t$. Here, $q=e$
The electric current is given by $i=\frac{2 \pi a_0}{v} e$.
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Thinking Process
The third line in Balmer series in the spectrum of hydrogen atom is $Hg$.
Ans.
$H _{\gamma}$ in Balmer series corresponds to transition $n=5$ to $n=2$. So, the electron in ground state i.e., from $n=1$ must first be placed in state $n=5$.
Energy required for the transition from $n=2$ to $n=5$ is given by
$=E_1-E_5=13.6-0.54=13.06 eV$
<Success>Sol. to be continued …</Success>
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$D =\mu_D ; m_e 1-\frac{m_e}{2 M} $
$ =m_e 1-\frac{m_e}{2 M} 1+\frac{m_e}{2 M} $
Reduced mass for
If for hydrogen deuterium, the wavelength is $\frac{\lambda_H}{\lambda_D}$
$ \frac{\lambda_D}{\lambda_H}=\frac{\mu_H}{\lambda_D} \simeq 1+\frac{m_e}{2 M} \simeq 1-\frac{1}{2 \times 1840} $
$ \lambda_D=\lambda_H \times(0.99973) $
On substituting the values, we have
Thus, lines are $1217.7 \AA , 1027.7 \AA, 974.04 \AA, 951.143 \AA$.
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Ques. Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in $1H _ 1$ and $1H _ 2$. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass $\mu$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mu$ = $\frac{m_eM}{(M + m)}$, where $M$ is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the $1^\text{st}$ line of the Lyman series in $1H _ 1$ and $1H _2$.
(Mass of $1H_1$ nucleus is $1.6725 \times 10^{-27} kg$, Mass of $1H _2$ nucleus is $3.3374 \times 10^{-27}$ $kg$, Mass of electron $=9.109 \times 10^{-31} kg$.)
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Thinking Process
The percentage difference in wavelength is given by
$100 \times \frac{\Delta \lambda}{\lambda_H}=\frac{\lambda_D-\lambda_H}{\lambda_H} \times 100, \text{ where signs are as usual. }$
Answer
The total energy of the electron in the $n$th states of the hydrogen like atom of atomic number $Z$ is given by
$E_n=-\frac{\mu Z^{2} e^{4}}{8 \varepsilon_0^{2} h^{2}} \frac{1}{n^{2}}$
<Success>Sol. to be continued …</Success>
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where signs are as usual and the $\mu$ that occurs in the Bohr formula is the reduced mass of electron and proton
Let $\mu_H$ be the reduced mass of hydrogen and $\mu_D$ that of Deutrium. Then, the frequency of the 1 st Lyman line in hydrogen is $h v_H=\frac{\mu_H e^{4}}{8 \varepsilon_0^{2} h^{2}} 1-\frac{1}{4}=\frac{\mu_H e^{4}}{8 \varepsilon_0^{2} h^{2}} \times \frac{3}{4}$.
Thus, the wavelength of the transition is $\lambda_H=\frac{3}{4} \frac{\mu_H e^{4}}{8 \varepsilon_0^{2} h^{3} C}$. The wavelength of the transition for the same line in Deutrium is $\lambda_D=\frac{3}{4} \frac{\mu_D e^{4}}{8 \varepsilon_0^{2} h^{3} c}$.
$\therefore \quad \Delta \lambda=\lambda_D-\lambda_H$
Hence, the percentage difference is
<Success>Sol. to be continued …</Success>
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$100 \times \frac{\Delta \lambda}{\lambda_H} =\frac{\lambda_D-\lambda_H}{\lambda_H} \times 100=\frac{\mu_D-\mu_H}{\mu_H} \times 100 $
$ =\frac{\frac{m_e M_D}{(m_e+M_D)}-\frac{m_e M_H}{(m_e-M_H)}}{\frac{m_e M_H}{(m_e+M_H.}} \times 100 $
$ =\frac{m_e+M_H}{m_e+M_D} \frac{M_D}{M_H}-1 \times 100 $
Since, $m_e \ll M_H \ll M_D$
$\frac{\Delta \lambda}{\lambda_H} \times 100 =\frac{M_H}{M_D} \times \frac{M_D}{M_H} \frac{1+\frac{m_e}{M_H}}{1+\frac{m_e}{M_D}}-1 \times 100 $
<Success>Sol. to be continued …</Success>
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$r_B^{\prime 4}=(0.51 \AA) . R^{3} \quad[R=10 \AA]$
$\therefore r_B^{\prime} \simeq(510)^{1 / 4} \AA<R$
$KE =\frac{1}{2} m v^{2}=\frac{m}{2} . \frac{h}{m^{2} r_B^{\prime 2}}=\frac{h}{2 m} . \frac{1}{r_B^{\prime 2}} $
$ =\frac{h^{2}}{2 m r_B^{2}} . \frac{r_B^{2}}{r_B^{\prime 2}}=(13.6 eV) \frac{(0.51)^{2}}{(510)^{1 / 2}}=\frac{3.54}{22.6}=0.16 eV $
$PE =+\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{r_B^{\prime 2}-3 R^{2}}{2 R^{3}} $
$ =+\frac{e^{2}}{4 \pi \varepsilon_0} . \frac{1}{r_B} . \frac{r_B^{\prime}(r_B^{2}-3 R^{2})}{R^{3}}=+(27.2 eV) \frac{0.51(\sqrt{510}-300)}{1000} $
$ =+(27.2 eV) . \frac{-141}{1000}=-3.83 eV $
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Thinking Process
As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in $Cr$, the energy states may be thought of as given by the Bohr model.
Ans.
The energy of the $n$th state $E_n=-Z^{2} R \frac{1}{n^{2}}$ where $R$ is the Rydberg constant and $Z=24$.
The energy released in a transition from 2 to 1 is $\Delta E=Z^{2} R \quad 1-\frac{1}{4}=\frac{3}{4} Z^{2} R$.
<Success>Sol. to be continued …</Success>
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Ques. The inverse square law in electrostatic is $|F|=\frac{e^{2}}{(4 \pi \varepsilon_0) r^{2}}$ for the force between an electron and a proton. The $\frac{1}{r}$ dependence of $|{F}|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$, force would be modified to $\bigg[F =\frac{e^{2}}{(4 \pi \varepsilon_0) r^{2}} \bigg[\frac{1}{r^{2}}+\frac{\lambda}{r}\bigg]e^{-\lambda}\bigg]$ where $\lambda=\frac{m_p c}{\hbar}$ and $\hbar=\frac{h}{2 \pi}$.
Estimate the change in the ground state energy of a H-atom if $m_p$ were $10^{-6}$ times the mass of an electron.
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The wavelength associated with is given by
where,
$\frac{\hbar}{m_p c} =\frac{\hbar c}{m_p c^{2}}=\frac{10^{-34} \times 3 \times 10^{8}}{0.8 \times 10^{-19}} $
$ \approx 4 \times 10^{-7} m \gg \text{ Bohr radius } $
$|{F}| =\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} \exp (-\lambda r) $
$\therefore \quad \lambda \ll \frac{1}{r_B}$ i.e., $\lambda r_B \ll 1$
<Success>Sol. to be continued …</Success>
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Also,
$ U(r)=-\frac{e^{2}}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda r)}{r} $
$ m v r=\hbar \quad \therefore \quad v=\frac{\hbar}{m r} $
$\frac{m v^{2}}{r}=\approx \frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} $
$ \therefore \quad \frac{\hbar^{2}}{m r^{3}}=\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r^{2}}+\frac{\lambda}{r} $
$ \therefore \quad \frac{\hbar^{2}}{m}=\frac{e^{2}}{4 \pi \varepsilon_0}[r+\pi r^{2}. $
<Success>Sol. to be continued …</Success>
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$\delta =\frac{-\lambda r_B^{2}}{1+2 \lambda r_B} \approx \lambda r_B^{2}(1-2 \lambda r_B)=-\lambda r_B^{2} $
Since, $\lambda r_B \ll 1$
$\therefore \quad V(r) =-\frac{e^{2}}{4 \pi \varepsilon_0} . \frac{\exp (-\lambda \delta-\lambda r_B)}{r_B+\delta} $
$\therefore \quad V(r) =-\frac{e^{2}}{4 \pi \varepsilon_0} \frac{1}{r_B} \quad 1-\frac{\delta}{r_B} .(1-\lambda r_B) $
$\cong(-27.2 eV) \text{ remains unchanged } $
<Success>Sol. to be continued …</Success>
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Ques. The Bohr model for the $H$-atom relies on the Coulomb’s law of electrostatics. Coulomb’s law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb’s law between two opposite charge $+q_1,-q_2$ is modified to
$|{F}| =\frac{q_1 q_2}{(4 \pi \varepsilon_0)} \frac{1}{r^{2}}, (r \geq R_0) $
$|F| =\frac{q_1 q_2}{4 \pi \varepsilon_0 R_0^2} \bigg[\frac{R_0}{r}\bigg]^r (r \leq R_0) $
Calculate in such a case, the ground state energy of a H-atom, if $\varepsilon=0.1$, and $R_0=1 \AA$.
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where,
$\frac{q_1 q_2}{4 \pi_0 \varepsilon_0} =(1.6 \times 10^{-19})^{2} \times 9 \times 10^{9} $ 
$ =23.04 \times 10^{-29} N m^{2} $
The electrostatic force of attraction between positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force.
$ =\frac{m v^{2}}{r} \text{ or } v^{2}=\frac{\Lambda R_0^{\delta}}{m r^{1+\delta}} $ 
$m v r =n h \cdot r=\frac{n \hbar}{m v}=\frac{n h}{m} \frac{m}{\Lambda R_0^{\delta}} r^{1 / 2} r^{1 / 2+\delta / 2} $
<Success>Sol. to be continued …</Success>
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[Applying Bohr’s second postulates]
Solving this for $r$, we get $\quad r_n={\frac{n^{2} \hbar^{2}}{m \wedge R_0^{\delta}}}^{\frac{1}{1-\delta}}$
where, $r_n$ is radius of $n$th orbit of electron. 
For $n=1$ and substituting the values of constant, we get
$ r_1 =\frac{\hbar^{2}}{m \wedge R_0^{\delta}} \frac{1}{1-\delta} $
$ r_1 =\frac{1.05^{2} \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 \times 10^{-28} \times 10^{+19}} $
<Success>Sol. to be continued …</Success>
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$=-\frac{\Lambda R_0^{\delta}}{1+\delta} \frac{1}{r^{1+\delta}}-\frac{1}{R_0^{1+\delta}}=-\frac{\Lambda}{1+\delta} \frac{R_0^{\delta}}{r^{1+\delta}}-\frac{1}{R_0}$
$PE =-\frac{\Lambda}{1+\delta} \frac{R_0^{\delta}}{r^{1+\delta}}-\frac{1}{R_0}+\frac{1+\delta}{R_0} $
$ PE =-\frac{\Lambda}{-0.9} \frac{R_0^{-1.9}}{r^{-0.9}}-\frac{1.9}{R_0} $
$ =\frac{2.3}{0.9} \times 10^{-18}[(0.8)^{0.9}-1.9] J=-17.3 eV $
Total energy is $(-17.3+5.9)=-11.4 eV$
This is the required TE of electron in ground state.
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