Physics · Atomic Physics

JEE Main 2026 — 5 April, Morning Shift — Question 10

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f) such that the ratio of radii $r_i : r_f = 16:4$ . The wavelength of photon emitted due to this transition is ___ nm. (Given Rydberg constant = 1.0973 × 10^7 /m)

Option A:
121
Option B:
242
Option C:
486
Correct
Option D:
974

Answer: C

Step-by-step solution

$r \propto n^2$ , so $n_i/n_f = \sqrt{16/4}=2$ , thus $n_i=4, n_f=2$ . $\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R\left(\frac{1}{4}-\frac{1}{16}\right)=R\cdot\frac{3}{16}$ , $\lambda = 16/(3R)=486$ nm.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

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Answer: C

Step-by-step solution

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