Physics - Modern Physics Question with Solution | TestHub

Energy for transition of electron from one orbit to other is given by-
$E_2-E_1=\left(13.6\right)Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=\frac{hc}{\lambda }$
$\Rightarrow \frac{1}{\lambda }=\frac{13.6}{hc}.Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Now as per question
$\frac{1}{{\lambda }_a}=\frac{13.6}{hc}.Z^2\left(\frac{1}{1^2}-\frac{1}{4^2}\right)$
and$\frac{1}{{\lambda }_e}=\frac{13.6}{hc}.Z^2\left(\frac{1}{m^2}-\frac{1}{4^2}\right)$
$\Rightarrow \frac{{\lambda }_e}{{\lambda }_a}=\frac{\left(1-\frac{1}{16}\right)}{\frac{1}{m^2}-\frac{1}{16}}=\frac{\left(15\right)m^2 }{16-m^2}$
But$\frac{{\lambda }_a}{{\lambda }_e}=\frac{1}{5}\left(\mathrm{g}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\right)$
$\Rightarrow \frac{16-m^2}{15 m^2}=\frac{1}{5}\Rightarrow 16-m^2=3m^2$
$\Rightarrow m^2=4$
$\Rightarrow m=2 \Rightarrow \left(\mathrm{C}\right)$is correct
Now,
$\frac{1}{{\lambda }_e}=\frac{13.6}{hc}.Z^2\left(\frac{1}{m^2}-\frac{1}{16}\right)$
$=\frac{13.6}{1242}\times 1\left(\frac{1}{4}-\frac{1}{16}\right) \left(\mathrm{a}\mathrm{s}\mathrm{ }hc=1242 \mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }z=1\right)$
$\Rightarrow {\lambda }_e=\frac{1242}{13.6}\times \frac{16}{3}=487 \mathrm{n}\mathrm{m}\mathrm{ }\Rightarrow \left(\mathrm{A}\right)$is incorrect
Now, kinetic energy of electron
$K\propto \frac{1}{n^2}$
$\Rightarrow \frac{K_2}{K_1}=\frac{1^2}{2^2}=\frac{1}{4}\Rightarrow \left(\mathrm{B}\right)$is correct
Also,$∆P=\frac{h}{\lambda }$
$\Rightarrow \frac{∆P_a}{∆P_e}=\frac{{\lambda }_e}{{\lambda }_a}=5$
$\Rightarrow \left(\mathrm{D}\right)$is incorrect.