Physics · Atomic Physics

JEE Main 2026 — 23 January, Morning Shift — Question 37

In hydrogen atom spectrum, ( $\mathrm{R} \rightarrow$ Rydberg's constant) A. the maximum wavelength of the radiation of Lyman series is $\frac{4}{3 R}$ B. the Balmer series lies in the visible region of the spectrum C. the minimum wavelength of the radiation of Paschen series is $\frac{9}{R}$ D. the minimum wavelength of Lyman series is $\frac{5}{4 R}$

Choose the correct answer from the options given below:

Option A:
B, D Only
Option B:
A, B and C Only
Correct
Option C:
A, B and D Only
Option D:
A, B Only

Answer: B

Step-by-step solution

$\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{4}\right)$ $\lambda=\frac{4}{3 R}$ $\frac{1}{\lambda^{\prime}}=\mathrm{R}\left(\frac{1}{9}\right)$ $\lambda^{\prime}=\frac{9}{\mathrm{R}}$

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

The de Broglie wavelength of an oxygen molecule at 27^circ C is x × 10^-12 m . The value of x is (take Planck's constant =6.63 × 10^-34 J .…

Answer: B

Step-by-step solution

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