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

Energy of electron in the 3^rd orbit of $$H{e^ + }$$ is
$$\eqalign{ & {E_3} = - 13.6 \times \frac{{{Z^2}}}{{{n^2}}}eV = - 13.6 \times \frac{4}{{{3^2}}}eV \cr & = - 13.6 \times \frac{4}{9} \times 1.6 \times {10^{ - 19}}\,J \cr} $$
From Bohr’s model,
$$\eqalign{ & {E_3} = - K{E_3} = - \frac{1}{2}{m_e}{v^2} \cr & \Rightarrow \frac{1}{2} \times 9.1 \times {10^{ - 31}} \times {v^2} = - 13.6 \times \frac{4}{9} \times 1.6 \times {10^{ - 19}} \cr & \Rightarrow {v^2} = \frac{{136 \times 16 \times 4 \times 2 \times {{10}^{ - 11}}}}{{9 \times 91}} \cr & {\text{or,}}\,\,v = 1.46 \times {10^6}\;m/s \cr} $$

Energy provided to the ground state electron
$$\eqalign{ & = \frac{{hc}}{\lambda } = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{975 \times {{10}^{ - 10}}}} = \frac{{6.6 \times 3}}{{975}} \times {10^{ - 16}} \cr & = 0.020 \times {10^{ - 16}} = 2 \times {10^{ - 18}}J \cr & = \frac{{20 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}eV = \frac{{20}}{{1.6}}eV = 12.75\,eV \cr} $$
It means the electron jumps to $$n = 4$$  from $$n = 1.$$  When electron will fall back, number of spectral lines emitted $$ = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6.$$

In Lyman series energy is released in U.V. region.
In Balmer series energy is released in Visible region.
In Paschen/Bracket/$$P$$ - fund series energy is released in I.R. region.

$$\eqalign{ & \frac{1}{{{\lambda _{3 - 1}}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{8R}}{9} \cr & \frac{1}{{{\lambda _{2 - 1}}}} = R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{{3R}}{4} \cr & \Rightarrow \frac{{{\lambda _{3 - 1}}}}{{{\lambda _{2 - 1}}}} = \frac{{27}}{{32}} \cr} $$

Frequency is given by
$$hv = - 13.6\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)$$
For transition from $$n = 6\,{\text{to}}\,n = 2,$$
$${v_1} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{6^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{2}{9} \times \left( {\frac{{13.6}}{h}} \right)$$
For transition from $$n = 2\,{\text{to}}\,n = 1,$$
$$\eqalign{ & {v_2} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{1^2}}}} \right) = \frac{3}{4} \times \left( {\frac{{13.6}}{h}} \right) \cr & \therefore {v_1} > {v_2} \cr} $$

The energy of the system of two atoms of diatomic molecule $$E = \frac{1}{2}I{\omega ^2}$$   where $$I = $$  moment of inertia
$$\omega = {\text{Angular velocity}} = \frac{L}{I},$$
$$L = $$  Angular momentum
$$I = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)$$
$$\eqalign{ & {\text{Thus,}}\,E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right){\omega ^2}\,......\left( {\text{i}} \right) \cr & E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{I^2}}} \cr & L = n\hbar \,\,\left( {{\text{According Bohr's Hypothesis}}} \right) \cr & E = \frac{1}{2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)\frac{{{L^2}}}{{{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}^2}}} \cr & E = \frac{1}{2}\frac{{{L^2}}}{{\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} = \frac{{{n^2}{h^2}}}{{8{\pi ^2}\left( {{m_1}r_1^2 + {m_2}r_2^2} \right)}} \cr & E = \frac{{\left( {{m_1} + {m_2}} \right){n^2}{h^2}}}{{8{\pi ^2}{r^2}{m_1}{m_2}}} \cr} $$

$$\eqalign{ & N \propto \frac{1}{{\frac{{{{\sin }^4}\theta }}{2}}};\frac{{{N_2}}}{{{N_1}}} = \frac{{{{\sin }^4}\left( {\frac{{{\theta _1}}}{2}} \right)}}{{{{\sin }^4}\left( {\frac{{{\theta _2}}}{2}} \right)}} \cr & {\text{or}}\,\,{N_2} = 5 \times {10^6} \times {\left( {\frac{1}{2}} \right)^4}{\left( {\frac{2}{{\sqrt 3 }}} \right)^4} = \frac{5}{9} \times {10^6} \cr} $$

$$\eqalign{ & \frac{1}{\lambda } = R{Z^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{3}{4}R{Z^2} \cr & {\text{or}}\,\,{Z^2} = \frac{4}{{3R\lambda }} = \frac{4}{{3 \times \left( {1.097 \times {{10}^7}} \right) \times \left( {0.76 \times {{10}^{ - 10}}} \right)}} = 1599.25 \cr & {\text{or}}\,\,{Z^2} = 1600 \cr & \therefore Z = 40. \cr} $$

Let initial number of nuclei in $$A$$ and $$B$$ is $${N_0}.$$
Number of nuclei of $$A$$ after time $$t$$ is
$${N_A} = {N_0}{e^{ - 8\lambda t}}\,......\left( {\text{i}} \right)$$
Similarly, number of nuclei of $$A$$ after time $$t$$ is
$${N_B} = {N_0}{e^{ - \lambda t}}\,......\left( {{\text{ii}}} \right)$$
It is given that $$\frac{{{N_A}}}{{{N_B}}} = \frac{1}{e}\,\,\left[ {\because {N_B} > {N_A}} \right]$$
Now, from Eqs. (i) and (ii)
$$\frac{{{e^{ - 8\lambda t}}}}{{{e^{ - \lambda t}}}} = \frac{1}{e}$$
Rearranging
$$\eqalign{ & \Rightarrow {e^{ - 1}} = {e^{ - 7\lambda t}} \Rightarrow 7\lambda t = 1 \cr & \Rightarrow {\text{Time}}\,t = \frac{1}{{7\lambda }} \cr} $$

$$\eqalign{ & \because B = \frac{{{\mu _0}I}}{{2r}}\,\,{\text{and}}\,\,I = \frac{e}{T} \cr & B = \frac{{{\mu _0}e}}{{2rT}}\,\,\left[ {r \propto {n^2},T \propto {n^5}} \right]; \cr & B \propto \frac{1}{{{n^5}}} \cr} $$