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JEE Main 2020 · ChemistryMultiple choiceSingle correctMediumCalculation

JEE Main 4 September 2020, Shift 2, Chemistry Q31

Question 31 of 75 in this shift, Chemistry question 6 of 25.

The shortest wavelength of H atom in the Lyman series is

\lambda_1

. The longest wavelength in the Balmer series of

\mathrm{He^+}

is :

(1) $\frac{36\lambda_1}{5}$
(2) $\frac{5\lambda_1}{9}$
(3) $\frac{9\lambda_1}{5}$ Official answer
(4) $\frac{27\lambda_1}{5}$

Official answer

Option 3

NTA final key (Sep 2020).

Chapter: Structure of Atom
Topic: Bohr's Atomic Theory
Idea tested: Rydberg Formula and Transitions

Same idea in other shifts

Asked 2× in all

4 Apr 2026, Shift 1 · Q52What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett…MediumSingle correct

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 108 of the 174 JEE Main shifts held from 2017 to 2026.