JEE Main 2020PhysicsAtomic PhysicsMediumNumerical

JEE Main 2020Atomic Physics Question with Solution

JEE Main 2020 (08 Jan Shift 2)

Question

The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A. The wavelength of the second member of the Balmer series (in nm) is_____________

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Show full solutionCorrect answer: 486
Correct answer
486

Step-by-step explanation

1λ=RZ21n12-1n22
1λ1=R12122-142=3R16
λ2λ1=2027
λ2=2027×6561 A=4860 A
=486 nm

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About this question

This is a previous-year question from JEE Main 2020, 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.