JEE Main 2022PhysicsAtomic PhysicsMediumNumerical

JEE Main 2022 — Atomic Physics Question with Solution

JEE Main 2022 (26 Jul Shift 1)

Question

In a hydrogen spectrum λ be the wavelength of first transition line of Lyman series. The wavelength difference will be "aλ" between the wavelength of 3rd transition line of Paschen series and that of 2nd  transition line of Balmer Series where a= _____.

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Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

Using Rydberg formula,

1λ=R1nf2-1ni2, where, R is Rydberg constant and n is quantum number.

For first line of Lyman series

1λ=R1-14=R34

⇒λ=43R      ...1

For 3rd  line of Paschen series

1λ3=R132-162=R9×34

For 2nd line of Balmer series

1λ2=R122-142=R4×34

Thus, aλ=λ3-λ2=12R-163R=203R

Putting equation 1

a43R=203R⇒a=5

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About this question

This is a previous-year question from JEE Main 2022, 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.