1. Use hydrogen energy levels

For hydrogen, $E_n=-\frac{13.6}{n^2}\text{ eV}$

The atom absorbs a photon and goes from $n=1$ to $n=4$. So absorbed energy is $\Delta E_a=E_4-E_1=-\frac{13.6}{16}-(-13.6)=13.6\left(1-\frac{1}{16}\right)$ $\Delta E_a=13.6\cdot \frac{15}{16}=12.75\text{ eV}$

Hence, $\lambda_a=\frac{hc}{\Delta E_a}=\frac{1242}{12.75}\text{ nm}$

Given $\frac{\lambda_a}{\lambda_e}=\frac{1}{5}$ so $\lambda_e=5\lambda_a$

Therefore emitted photon energy is $E_e=\frac{hc}{\lambda_e}=\frac{1}{5}\frac{hc}{\lambda_a}=\frac{1}{5}\Delta E_a=\frac{12.75}{5}=2.55\text{ eV}$

2. Find the final state $m$

Emission occurs from $n=4$ to $n=m$. Thus, $E_e=13.6\left(\frac{1}{m^2}-\frac{1}{16}\right)$

Set this equal to $2.55$ eV: $13.6\left(\frac{1}{m^2}-\frac{1}{16}\right)=2.55$

Divide by $13.6$: $\frac{1}{m^2}-\frac{1}{16}=\frac{2.55}{13.6}=0.1875=\frac{3}{16}$

So, $\frac{1}{m^2}=\frac{3}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$ $m^2=4 \Rightarrow m=2$

So option B is correct.

3. Check option A: ratio of kinetic energies

In Bohr model, kinetic energy in level $n$ is $K_n=\frac{13.6}{n^2}\text{ eV}$

Thus, $\frac{K_m}{K_1}=\frac{1/m^2}{1}$ Since $m=2$, $\frac{K_2}{K_1}=\frac{1}{4}$

So option A is correct.

4. Check option C: momentum change ratio

Photon momentum magnitude is $p=\frac{h}{\lambda}$

Since atom is initially free, momentum change due to absorption equals absorbed photon momentum: $\Delta p_a=\frac{h}{\lambda_a}$

Similarly for emission, $\Delta p_e=\frac{h}{\lambda_e}$

Hence, $\frac{\Delta p_a}{\Delta p_e}=\frac{h/\lambda_a}{h/\lambda_e}=\frac{\lambda_e}{\lambda_a}=5$

So it is not $\frac{1}{2}$. Thus option C is incorrect.

5. Check option D: emitted wavelength

We already have $\lambda_e=5\lambda_a$ with $\lambda_a=\frac{1242}{12.75}\approx 97.41\text{ nm}$

Therefore, $\lambda_e\approx 5\times 97.41=487.05\text{ nm}$

So option D ($418$ nm) is incorrect.

(Also, for $4\to 2$ transition in hydrogen, Balmer line is about $486$ nm, consistent with this.)

6. Final answer

Correct options are: $\boxed{A,\ B}$

This matches the stored correct answer.