A free hydrogen atom after absorbing a photon of wavelength  ${\lambda }_a$ gets excited from the state $n=1$to the state  $n=4$. Immediately after that, the electron jumps to  $n=m$ state by emitting a photon of wavelength  ${\lambda }_e$. Let the change in momentum of atom due to the absorption and the emission are $\Delta p_a$  and  $\Delta p_e$ , respectively. If  ${\lambda }_a/{\lambda }_e=1/5$. Which of the option(s) is/are correct? [Use $hc=1242eVnm;1nm={10}^{-9}m,h$  and  $c$  are Plank’s constant and speed of light, respectively]

Energy absorbed by electron in transition from n = 1 to n = 4 is given as
$\frac{hc}{{\lambda }_a}=13.6[\frac{1}{1}-\frac{1}{4^2}] \mathrm{...}\left(1\right)$ 
Energy released by electron in transition from n = 4 to n = m is given as
 $$\begin{gathered} \frac{{\lambda }_a}{{\lambda }_e}=\frac{[\frac{1}{m^2}-\frac{1}{16}]}{[1-\frac{1}{16}]}=\frac{1}{5} \mathrm{...}\left(1\right) \\ \Rightarrow \frac{1}{m^2}-\frac{1}{16}=\frac{15}{16}\times \frac{1}{5} \\ \Rightarrow \frac{1}{m^2}-\frac{1}{16}=\frac{3}{16} \\ \Rightarrow \frac{1}{m^2}=\frac{3}{16}+\frac{1}{16} \\ \Rightarrow m=2 \end{gathered}$$
Hence option (C) is correct.
From equation (2), we have
$$\begin{gathered} \frac{hc}{{\lambda }_e}=13.6[\frac{1}{2^2}-\frac{1}{4^2}]=13.6\times \frac{3}{16}eV \\ \Rightarrow {\lambda }_e=\frac{12400\times 16}{136\times 3}A \\ \Rightarrow {\lambda }_e\approx \text{4862}\stackrel{\text{0}}{\text{A}} \end{gathered}$$ 
Hence option (A ) is NOT correct.
For kinetic energy of electron in nth orbit, we use
$K{E}_n\propto \frac{z^2}{n^2}$ 
$\Rightarrow \frac{K{E}_2}{K{E}_1}=\frac{1}{4}$ 
Hence option (B) is correct.
Momentum of photon is given as 
and  $\Delta P_e=\frac{h}{{\lambda }_e}$
 $\Rightarrow \frac{\Delta P_a}{\Delta P_e}=\frac{{\lambda }_e}{{\lambda }_a}$
Hence option (D) is NOT correct.