The energy of the first (lowest) Balmer line of H -atom is x\, J . The energy (in J ) of the second Balmer line of H -atom is:

Chemistry · Structure of Atom

JEE Main 2026 — 22 January, Evening Shift — Question 57

The energy of the first (lowest) Balmer line of $\mathrm{H}$ -atom is $x\,\mathrm{J}$ . The energy (in $\mathrm{J}$ ) of the second Balmer line of $\mathrm{H}$ -atom is:

Option A:
$x^{2}$
Option B:
$\frac{x}{1.35}$
Option C:
$2 x$
Option D:
$1.35x$
Correct

Answer: D

Step-by-step solution

Balmer series ends at $n = 2$ .

First Balmer line: $3 \to 2$

= \frac{5}{36}$$ Second Balmer line: $4 \to 2$ $$E_2 \propto \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16}$$ Ratio, we have $$\frac{E_2}{E_1} = \frac{3/16}{5/36} = 1.35$$

E_2 = 1.35x

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Structure of Atom
Topic: Analysis of Spectra of H-like species