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JEE Main 2026 · ChemistryMultiple choiceSingle correctMediumMulti-step

JEE Main 6 April 2026, Shift 1, Chemistry Q52

Question 52 of 75 in this shift, Chemistry question 2 of 25, Section A.

If shortest wavelength of hydrogen atom in Lyman series is

x

, then longest wavelength in Balmer series of

\mathrm{He^+}

is:

(1) $\frac{9x}{5}$ Official answer
(2) $\frac{36x}{5}$
(3) $\frac{x}{4}$
(4) $\frac{5x}{9}$

Official answer

Option 1

NTA final key.

Chapter: Structure of Atom
Topic: Bohr's Atomic Theory
Idea tested: Longest and Shortest Wavelengths in a Series

Same idea in other shifts

Asked 2× in all

8 Jan 2020, Shift 1 · Q30For the Balmer series in the spectrum of H atom, $\bar{\nu} = R_H\left\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\right\}$ , the correct statements…MediumOther

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.