Chemistry · Structure of Atom

JEE Main 2026 — 28 January, Morning Shift — Question 55

The wave numbers of three spectral lines of H-atom are considered. Identify the set of spectral lines belonging to Balmer series. ( $\mathrm{R}=$ Rydberg constant)

Option A:
$\frac{5 \mathrm{R}}{36}, \frac{3 \mathrm{R}}{16}, \frac{21 \mathrm{R}}{100}$
Correct
Option B:
$\frac{5 \mathrm{R}}{36}, \frac{8 \mathrm{R}}{9}, \frac{15 \mathrm{R}}{16}$
Option C:
$\frac{7 \mathrm{R}}{144}, \frac{3 \mathrm{R}}{16}, \frac{16 \mathrm{R}}{255}$
Option D:
$\frac{3 \mathrm{R}}{4}, \frac{3 \mathrm{R}}{16}, \frac{7 \mathrm{R}}{144}$

Answer: A

Step-by-step solution

Balmer series corresponds to $n_1 = 2$

Wave number relation, we have

\bar{\nu} = R \left(\frac{1}{4} - \frac{1}{n_2^2}\right)

For $n_2 = 3$ , $\bar{\nu} = \frac{5R}{36}$

For $n_2 = 4$ , $\bar{\nu} = \frac{3R}{16}$

For $n_2 = 5$ , $\bar{\nu} = \frac{21R}{100}$

These all match option A.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Analysis of Spectra of H-like species

Figure 1. Electron probability density for 2s orbital Figure 2. Wave function for 2s orbital Which of the following point in Figure 2 most…

Answer: A

Step-by-step solution

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