The ratio of momentum of the photons of the 1 st and 2 nd line of Balmer series of Hydrogen atoms is α/β . The possible values of α and β are :

Physics · Atomic Physics

JEE Main 2026 — 6 April, Morning Shift — Question 14

The ratio of momentum of the photons of the $1^{\mathrm{st}}$ and $2^{\mathrm{nd}}$ line of Balmer series of Hydrogen atoms is $\alpha/\beta$ . The possible values of $\alpha$ and $\beta$ are :

Option A:
27 and 20
Option B:
3 and 16
Option C:
5 and 36
Option D:
20 and 27
Correct

Answer: D

Step-by-step solution

For Balmer series: $1/\lambda = R(1/2^2 - 1/n^2)$ . First line (n=3): $1/\lambda_1 = R(1/4-1/9)=5R/36$ . Second line (n=4): $1/\lambda_2 = R(1/4-1/16)=3R/16$ . Momentum $p = h/\lambda$ , so $p_1/p_2 = \lambda_2/\lambda_1 = (16/(3R)) / (36/(5R)) = (16/3)*(5/36)=80/108=20/27$ . Thus $\alpha=20,\beta=27$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum