QJEEAdv19Phe381: JEE Advanced 2019 Physics - Class 12 JEEAdv19Phe381 Solution: A free hydrogen atom after absorbing a photon of wavelength a…

A hydrogen atom in its ground state (n = 1) absorbs a photon with wavelength $\lambda_a$ and gets excited to the n = 4 state. Subsequently, the electron transitions to the n = m state, emitting a photon with wavelength $\lambda_e$ . Given that ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$ , let's analyze the options systematically.

Calculation for the Wavelengths:

The energy difference between levels in a hydrogen atom can be given by the Rydberg formula for hydrogen:

$E_n = - \frac{{13.6 \ eV}}{{n^2}}$

The energy of the absorbed photon that excites the electron from n = 1 to n = 4 could be expressed as:

$E_a = E_4 - E_1 = - \frac{{13.6 \ eV}}{{4^2}} - (- \frac{{13.6 \ eV}}{{1^2}}) = - \frac{{13.6 \ eV}}{{16}} + 13.6 \ eV = 13.6 \ eV (1 - \frac{1}{16}) = 13.6 \ eV \cdot \frac{15}{16} = 12.75 \ eV$

Wavelength $\lambda_a$ associated with this energy is:

$\lambda_a = \frac{{hc}}{{E_a}} = \frac{{1242 \ nm \cdot eV}}{{12.75 \ eV}} = 97.4 \ nm$

Given ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$ , we get:

$\lambda_e = 5 \lambda_a = 5 \cdot 97.4 \ nm = 487 \ nm$

Option A: The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is ${1 \over 4}$

The kinetic energy of an electron in a hydrogen atom in the n^th state is proportional to $- \frac{{1}}{{n^2}}$ . Therefore, the ratio of kinetic energy in state m to that in state n can be written as:

$\frac{{E_m}}{{E_1}} = \frac{{- \frac{{13.6}}{{m^2}} eV}}{{- 13.6 \ eV}} = \frac{1}{{m^2}}$

According to the question, the ratio is $\frac{1}{4}$ :

$\frac{1}{{m^2}} = \frac{1}{4} \Rightarrow m = 2$

Therefore, Option A is correct if m = 2.

Option B: m = 2

Based on the energy calculations above, we find m = 2. Therefore, Option B is correct.

Option C: ${{\Delta {p_a}} \over {\Delta {p_e}}} = {1 \over 2}$

The change in momentum for each photon is given by:

$\Delta p = \frac{h}{\lambda}$

So, the ratio of the momentum changes is:

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

This shows the given option C (i.e., $\frac{1}{2}$ ) is incorrect. The correct ratio is 5.

Option D: $\lambda_e = 418 \ nm$

From our previous calculations, we found $\lambda_e = 487 \ nm$ , which does not match 418 nm. Option D is incorrect.

Hence, the correct options are A and B.