JEE Main 2019PhysicsAtomic PhysicsEasyMCQ

JEE Main 2019 — Atomic Physics Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

Taking the wavelength of first Balmer line in hydrogen spectrum (n=3  to n=2) as 660 nm , the wavelength of the 2nd Balmer line (n = 4  to  n = 2) will be :

Choose an option

Show full solutionCorrect option: B
Correct answer
B488.9 nm

Step-by-step explanation

From Rydberg’s equation,
1λ=R1n2-1n22
1660×10-9=R122-132
1λ=R122-142
λ600×10-9=59×4×16×412
λ=488.9 nm

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About this question

This is a previous-year question from JEE Main 2019, covering the Atomic Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.