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

$$ - \frac{{dU}}{{dr}} = F\,\,\left( {{\text{conservative force field}}} \right)$$
$$ \Rightarrow F = \frac{{ - K}}{r}$$   provides the centrifugal force for circular motion of electron.
$$\frac{{m{v^2}}}{r} = \frac{K}{r} \Rightarrow r = \frac{{nh}}{{2\pi \sqrt {mK} }}$$
$$K.E.$$  of electron $$ = \frac{1}{2}m{v^2} = \frac{1}{2}K$$
$$P.E.$$  of electron $$ = K\ln r$$
$$\eqalign{ & E\left( n \right) = {\text{Total}}\,{\text{energy}} = K.E. + P.E. \cr & = \frac{1}{2}K + K\ln \,r = \frac{K}{2}\left[ {1 + \log \frac{{{n^2}{h^2}}}{{4{\pi ^3}mk}}} \right] \cr} $$
Required ratio $$ = \frac{{E\left( 2 \right) - E\left( 1 \right)}}{{E\left( 4 \right) - E\left( 2 \right)}} = 1$$

The radius of $$n$$^th Bohr's orbit of hydrogen and hydrogen like atom
$$\eqalign{ & {r_n} = \frac{{{\varepsilon _0}{n^2}{h^2}}}{{\pi m{e^2}Z}} \cr & \therefore {r_n} = \frac{{{n^2}{a_0}}}{Z}\,\,{\text{or}}\,\,{r_n} \propto \frac{{{n^2}}}{Z} \cr} $$
For ground state, $$n =1$$
Atomic number, $$Z = 1$$
For first excited state, $$n = 2$$
$$\therefore \frac{{{r_2}}}{{{r_1}}} = {\left( {\frac{2}{1}} \right)^2} = 4\,\,{\text{or}}\,\,{r_2} = 4{r_1}$$
Therefore, radius of first excited state is 4 times than that of ground state radius in $$H$$-atom.

Total energy of electron for hydrogen like atom is given by
$$\eqalign{ & {E_n} = - \frac{{13.6{Z^2}}}{{{n^2}}} \cr & \therefore {E_3} = - \frac{{13.6}}{{{3^2}}}eV\,\,\left[ {Z = 1,n = 3} \right] \cr & = - 1.51\,eV \cr & {E_4} = - \frac{{13.6}}{{{4^2}}} = - 0.85\,eV \cr & \therefore {E_4} - {E_3} = 1.51 - 0.85 = 0.66\,eV \cr} $$

$$\eqalign{ & \Delta E = hv \cr & v = \frac{{\Delta E}}{h} = k\left[ {\frac{1}{{{{\left( {n - 1} \right)}^2}}} - \frac{1}{{{n^2}}}} \right] = \frac{{k\left( {2n - 1} \right)}}{{{n^2}{{\left( {n - 1} \right)}^2}}} \cr & \approx \frac{{2k}}{{{n^3}}}\,\,{\text{or,}}\,\,v \propto \frac{1}{{{n^3}}} \cr} $$

Using Bohr's postulate for radiation of spectral line, we have
Radiation of wavelength from $$C$$ to $$B$$
$${E_C} - {E_B} = \frac{{hc}}{{{\lambda _1}}}\,.......\left( {\text{i}} \right)$$
Radiation of wavelength from $$B$$ to $$A$$
$${E_B} - {E_A} = \frac{{hc}}{{{\lambda _2}}}\,.......\left( {{\text{ii}}} \right)$$
Radiation of wavelength from $$C$$ to $$A$$
$${E_C} - {E_A} = \frac{{hc}}{{{\lambda _3}}}\,.......\left( {{\text{iii}}} \right)$$
Also, $$\left( {{E_C} - {E_A}} \right) = \left( {{E_C} - {E_B}} \right) + \left( {{E_B} - {E_A}} \right)$$
$$\eqalign{ & \therefore \frac{{hc}}{{{\lambda _3}}} = \frac{{hc}}{{{\lambda _1}}} + \frac{{hc}}{{{\lambda _2}}}\, \cr & {\text{or}}\,\,\frac{1}{{{\lambda _3}}} = \frac{1}{{{\lambda _1}}} + \frac{1}{{{\lambda _2}}} \cr & \Rightarrow {\lambda _3} = \frac{{{\lambda _1}{\lambda _2}}}{{{\lambda _1} + {\lambda _2}}} \cr} $$

Excess energy of $${e^ - }$$ appears as photon.
From Rydberg’s formula,
$$\eqalign{ & \frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right) = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}} \cr & \frac{1}{{\lambda '}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right) = \frac{{7R}}{{144}} \cr & \frac{{\frac{1}{\lambda }}}{{\frac{1}{{\lambda '}}}} = \frac{{5R}}{{36}} \div \frac{{7R}}{{144}} \cr & \Rightarrow \frac{{\lambda '}}{\lambda } = \frac{{5R}}{{36}} \times \frac{{144}}{{7R}} = \frac{{20}}{7} \cr & \Rightarrow \lambda ' = \frac{{20}}{7}\lambda \cr} $$

Energy of photon corresponding to first line of Balmer series $$ = \left( {13.6} \right){\left( 2 \right)^2}\left[ {\frac{1}{4} - \frac{1}{9}} \right]$$
Energy need to eject electron from $$n = 2$$  level in $$H$$ atom $$ = \left( {13.6} \right)\left( {\frac{1}{4}} \right)$$
So, required kinetic energy
$$\eqalign{ & = \left( {13.6} \right)\left[ {\left( {\frac{1}{4} - \frac{1}{9}} \right) - \left( {\frac{1}{4}} \right)} \right]eV \cr & = 13.6 \times \left( {\frac{{11}}{{36}}} \right) \cr & = 4.155\,eV \cr} $$

In the n^th orbit, let $${r_n}$$ be the radius and $${v_n}$$ be the speed of electron.
Time period, $${T_n} = \frac{{2\pi {r_n}}}{{{v_n}}} \propto \frac{{{r_n}}}{{{v_n}}}$$
Now $${r_n} \propto {n^2};{v_n} \propto \frac{1}{n}$$
$$\therefore \frac{{{r_n}}}{{{v_n}}} \propto {n^3}\,\,{\text{or}}\,\,{T_n} \propto {n^3}$$
Here $$8 = {\left( {\frac{{{n_1}}}{{{n_2}}}} \right)^3}\,\,{\text{or}}\,\,\frac{{{n_1}}}{{{n_2}}} = 2$$
i.e., $${n_1} = 2{n_2}$$

$$\eqalign{ & \frac{\nu }{x} = \frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] \cr & \left[ {\because n > > 1} \right] \cr & \nu = RC{Z^2}\left[ {\frac{1}{{{{\left( {n - 1} \right)}^2}}} - \frac{1}{{{n^2}}}} \right] \cr & RC{Z^2}\left[ {\frac{{2n - 1}}{{{n^2}{{\left( {n - 1} \right)}^2}}}} \right] \cr} $$

Centripetal force = Coulombian force
$$\eqalign{ & \frac{{m{v^2}}}{{{a_0}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{e \times e}}{{a_0^2}} \cr & \Rightarrow {v^2} = \frac{{{e^2}}}{{4\pi {\varepsilon _0}.{a_0}.m}} \cr & \Rightarrow v = \frac{e}{{\sqrt {4\pi {\varepsilon _0}.{a_0}.m} }} \cr} $$