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

We know that
$$\frac{1}{\lambda } = R{Z^2}\left[ {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right] \Rightarrow \frac{1}{\lambda } \propto {Z^2}$$
$$\lambda $$ is shortest when $$\frac{1}{\lambda }$$ is largest i.e., when $$Z$$ has a higher value. $$Z$$ is highest for lithium.

$$E$$ is minimum when $${n_1} = 1\,\& \,{n_2} = 2$$
$$ \Rightarrow {E_{\min }} = 13.6\left( {\frac{1}{1} - \frac{1}{4}} \right)eV = 13.6 \times \frac{3}{4}eV$$
$$E$$ is maximum when $${n_1} = 1\,\& \,{n_2} = \infty $$    (the atom is ionized, that is known as ionization energy)
$$\eqalign{ & \Rightarrow {E_{\max }} = 13.6\left( {1 - \frac{1}{\infty }} \right) = 13.6\,eV \cr & \therefore \frac{{{E_{\min }}}}{{{E_{\max }}}} = \frac{3}{4} \cr & \Rightarrow \frac{{\frac{{hc}}{{{\lambda _{\max }}}}}}{{\frac{{hc}}{{{\lambda _{\min }}}}}} = \frac{3}{4} \cr & \Rightarrow \frac{{{\lambda _{\min }}}}{{{\lambda _{\max }}}} = \frac{3}{4} \cr} $$

$$\eqalign{ & b = \frac{{Z{e^2}\cot \left( {\frac{\theta }{2}} \right)}}{{4\pi { \in _0}{k_i}}} = 0 \cr & \Rightarrow \cot \left( {\frac{\theta }{2}} \right) = 0 \cr & \Rightarrow \left( {\frac{\theta }{2}} \right) = {90^ \circ }\,\,{\text{or}}\,\,\theta = {180^ \circ } \cr} $$

$$\eqalign{ & E = hf = 13.6\,{Z^2}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right) \cr & \frac{{{f_2}}}{{{f_1}}} = \frac{{{3^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)}}{{{1^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right)}} \cr & \Rightarrow {f_2} = \frac{{243}}{{32}}f. \cr} $$