A free hydrogen atom after absorbing a photon of wavelength λ 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 λ e . Let the change in momentum of atom due to the absorption and the emission are Δ p a and Δ p e . respectively. If frac λ a λ e =1/5 , which of the option(s) is /are correct ? [Use hc =1242 eV nm ; 1 nm =10 -9 m , h and c are Planck's constant and speed of light, respectively]

The ratio of kinetic energy of the electron in the state n=m to the state n=1 is 1/4

Physics · Atomic Physics

JEE Advanced 2019 — Paper 2 — Question 7

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 $\Delta \mathrm{p}_{\mathrm{a}}$ and $\Delta \mathrm{p}_{\mathrm{e}}$ . respectively. If

$\frac{\lambda_{a}}{\lambda_{e}}=\frac{1}{5}$ , which of the option(s) is /are correct ? [Use hc $=1242 \mathrm{eV} \mathrm{nm} ; 1 \mathrm{~nm}=10^{-9} \mathrm{~m}$ , h and c are

Planck's constant and speed of light, respectively]

Option A:
$\frac{\Delta \mathrm{p}_{\mathrm{a}}}{\Delta \mathrm{p}_{\mathrm{e}}}=\frac{1}{2}$
Option B:
The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $\frac{1}{4}$
Correct
Option C:
$\lambda_{e}=418 \mathrm{~nm}$
Option D:
$\mathrm{m}=2$
Correct

Answer: B, D

Step-by-step solution

The energy level diagram is shown in the figure

It is given : $\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{e}}}=\frac{1}{5}$

or, $\frac{\frac{1}{-\mathrm{m}^{2}}-\frac{1}{4^{2}}-}{\frac{1}{-1}-\frac{1}{4^{2}}-}=\frac{\frac{1}{\lambda_{\mathrm{e}}}}{\frac{1}{\lambda_{\mathrm{a}}}}=\frac{1}{5}$

$\Rightarrow \mathrm{m}=2$

clearly $\frac{\Delta \mathrm{p}_{\mathrm{a}}}{\Delta \mathrm{p}_{\mathrm{e}}} \neq \frac{1}{2}$

The ratio of the kinetic energies is also the ratio of the corresponding total energies $=\frac{1}{4}$

$\therefore$ correct options are B, D

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency Omega such that frac4 pi…

Answer: B, D

Step-by-step solution

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