Question 101: Atoms & Nuclei

Question

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 be $$\Delta {p_a}$$ and $$\Delta {p_e}$$, respectively. If $${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$$, which of the option(s) is/are correct? [Use hc = 1242 eVnm; 1 nm = 10-9 m, h and c are Planck's constant and speed of light in vacuum, respectively]

Accepted Answer

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.

Answer

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

Answer

m = 2

Answer

$${{\Delta {p_a}} \over {\Delta {p_e}}} = {1 \over 2}$$

Answer

$$\lambda $$e = 418 nm

Atoms & Nuclei

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