Learn

Friday, 9 October

JEE Main 2020 · ChemistryMultiple choiceOtherMediumConceptual

JEE Main 8 January 2020, Shift 1, Chemistry Q30

Question 30 of 75 in this shift, Chemistry question 5 of 25, Section A.

For the Balmer series in the spectrum of H atom, νˉ=RH{1n12−1n22}\bar{\nu} = R_H\left\{\frac{1}{n_1^2}-\frac{1}{n_2^2}\right\}, the correct statements among (I) to (IV) are : (I) As wavelength decreases, the lines in the series converge (II) The integer n1n_1 is equal to 2 (III) The lines of longest wavelength corresponds to n2=3n_2 = 3 (IV) The ionization energy of hydrogen can be calculated from wave number of these lines
  1. (1)(I), (II), (III)Official answer
  2. (2)(II), (III), (IV)
  3. (3)(I), (III), (IV)
  4. (4)(I), (II), (IV)

Official answer

Option 1

NTA final key (Jan 2020).

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
  1. 6 Apr 2026, Shift 1 · Q52If shortest wavelength of hydrogen atom in Lyman series is xx, then longest wavelength in Balmer series of He+\mathrm{He^+} is: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 42 of the 174 JEE Main shifts held from 2017 to 2026.