Atoms And Nuclei MCQ Questions & Answers in Modern Physics | Physics

NOTE:
As the electron comes nearer to the nucleus the potential energy decreases
$$\left( {\because \frac{{ - k.Z{e^2}}}{r} = P.E.\,{\text{and}}\,r{\text{ decreases}}} \right)$$
The $$K.E.$$  will increase $$\left[ {\because K.E. = \frac{1}{2}\left| {P.E.} \right| = \frac{1}{2}\frac{{kZ{e^2}}}{r}} \right]$$
The total energy decreases $$\left[ {T.E. = - \frac{1}{2}\frac{{kZ{e^2}}}{r}} \right]$$

Energy $$E$$ of an atom with principal quantum number $$n$$ is given by $$E = \frac{{ - 13.6}}{{{n^2}}}{Z^2}$$    for first excited state $$n = 2$$  and for $$H{e^ + },Z = 2$$
So, $$E = \frac{{ - 13.6 \times {{\left( 2 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = - 13.6\,eV$$

KEY CONCEPT: $$R = {R_0}{\left( A \right)^{\frac{1}{3}}}$$
$$\eqalign{ & \therefore \frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{{A_1}}}{{{A_2}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{{27}}{{125}}} \right)^{\frac{1}{3}}} = \frac{3}{5} \cr & {R_2} = \frac{5}{3} \times 3.6 = 6fermi \cr} $$

$$\Delta E = {E_0}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right]$$
here (I^st) transition represent absorption not emission. So, (C) is correct option.

KEY CONCEPT:
$${E_n} = - 13.6\frac{{\left( {{Z^2}} \right)}}{{\left( {{n^2}} \right)}}eV$$
Therefore, ground state energy of doubly ionized lithium atom $$\left( {Z = 3,n = 1} \right)$$    will be
$${E_1} = \left( { - 13.6} \right)\frac{{{{\left( 3 \right)}^2}}}{{{{\left( 1 \right)}^2}}} = - 122.4eV$$
∴ Ionization energy of an electron in ground state of doubly ionized lithium atom will be $$122.4\,eV.$$

From Bohr's postulate, for any permitted (stationary orbit).
Angular momentum of electron revolving in an orbit is constant
i.e. $$mvr = \frac{{nh}}{{2\pi }}\,\,{\text{or}}\,\,v = \frac{{nh}}{{2\pi mr}}\,......\left( {\text{i}} \right)$$
Also, $$\frac{{m{v^2}}}{r} = \frac{{Z{e^2}}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{kZ{e^2}}}{{{r^2}}}\,......\left( {{\text{ii}}} \right)\,\,\left( {{\text{where,}}\,k = \frac{1}{{4\pi {\varepsilon _0}}}} \right)$$
Symbols have their usual meaning.
From Eqs. (i) and (ii),
$$r = \frac{{{n^2}{h^2}}}{{4{\pi ^2}mkZ{e^2}}}$$
For hydrogen atom,
$$\eqalign{ & Z = 1 \cr & \therefore r = \frac{{{n^2}{h^2}}}{{4{\pi ^2}mk{e^2}}} \cr & \Rightarrow {r_n} \propto {n^2} \cr & \therefore {a_2} = 4{a_0} \cr} $$
NOTE
For solving the problem, dependence of radius of $$n$$^th orbit of hydrogen like atom must be kept in mind i.e. $${r_n} \propto \frac{{{n^2}}}{Z}$$   [where, $$n$$ = $$n$$^th orbit and $$Z$$ = atomic number]

The energy of hydrogen atom when the electron revolves in $$n$$^th orbit is given by
$$E = \frac{{ - 13.6\,{Z^2}}}{{{n^2}}}eV\,\,\left[ {Z = 1} \right]$$
In the ground state; $$n = 1$$
$$E = \frac{{ - 13.6}}{{{1^2}}} = - 13.6\,eV$$
For $$n = 2,$$
$$E = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\,eV$$
So, kinetic energy of electron in the first excited state (i.e. for $$n = 2$$ ) is
$$KE = - E = - \left( { - 3.4} \right) = 3.4\,eV$$

$$qVB = \frac{{m{v^2}}}{r}\,\,{\text{and}}\,\,\frac{{nh}}{{2\pi }} = mvr$$
Multiplying equation (i) and (ii),
$$\frac{{qBnh}}{{2\pi }} = {m^2}{v^2}$$
Now multiplying both sides by $$\frac{1}{{2m}},$$
$$n\frac{{qBh}}{{4\pi m}} = \frac{1}{2}m{v^2}\,\,{\text{i}}{\text{.e}}{\text{.}}\,\,KE = n\left[ {\frac{{qBh}}{{4\pi m}}} \right]$$

Given potential energy between electron and proton
$$\eqalign{ & = e{V_0}\log \frac{r}{{{r_0}}}\quad \left[ {\because \left| U \right|\, = eV} \right] \cr & \therefore \left| F \right| = \frac{d}{{dr}}\left[ {e{V_0}{{\log }_e}\frac{r}{{{r_0}}}} \right] = \frac{{e{V_0}}}{{{r_0}}} \times \frac{1}{r} \cr} $$
But this force acts as centripetal force
$$\therefore \frac{{m{v^2}}}{r} = \frac{{e{V_0}}}{{r{r_0}}} \Rightarrow m{v^2} = \frac{{e{V_0}}}{{{r_0}}}\,......\left( {\text{i}} \right)$$
By Bohr’s postulate, $$mvr = \frac{{nh}}{{2\pi }}\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii),
$$\eqalign{ & \frac{{{m^2}{v^2}{r^2}}}{{m{v^2}}} = \frac{{{n^2}{h^2}{r_0}}}{{4{\pi ^2} \times {V_0}e}} \cr & \Rightarrow {r^2} = \frac{{{n^2}{h^2}{r_0}}}{{4\pi {V_0}me}} \Rightarrow r \propto n \cr} $$

KEY CONCEPT:
According to Doppler’s effect of light, the wavelength shift is given by
$$\eqalign{ & \Delta \lambda = \frac{v}{c} \times \lambda \cr & \Rightarrow v = \frac{{\Delta \lambda \times c}}{\lambda } = \frac{{\left( {706 - 656} \right)}}{{656}} \times 3 \times {10^8} \approx 2 \times {10^7}m/s \cr} $$