JEE Main 2020PhysicsAtomic PhysicsMediumNumerical

JEE Main 2020 — Atomic Physics Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 305 A∘. The corresponding difference for the Paschan series in A∘ is:________

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Show full solutionCorrect answer: 10553.14
Correct answer
10553.14

Step-by-step explanation

1λmin=R1=R ; n=∞ to 1

1λmax=R1−14=3R4 ; n=2 to 1

⇒ λmax−λmin=43R−1R

304=13R                                              ......a

Paschen : Yλ'min=R19

and Yλ'max=R19−116=7R16×9

λ'max−λ'min=16×97R−9R=817R       ...b

ba=x3×304=817

⇒ x=10553.14

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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.