JEE Main 2024PhysicsAtomic PhysicsHardNumerical

JEE Main 2024 — Atomic Physics Question with Solution

JEE Main 2024 (29 Jan Shift 1)

Question

When a hydrogen atom going from n=2 to n=1 emits a photon, its recoil speed is x5 m s-1. Where x=_______. (Use: mass of hydrogen atom =1.6×10-27 kg, charge of electron e=1.6×10-19 C)

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Show full solutionCorrect answer: 17
Correct answer
17

Step-by-step explanation

According to the above diagram, the difference in energy between the two states is given by

ΔE=-3.4 eV--13.6 eV=10.2 eV

The formula to calculate the recoil speed is given by

v=∆Emc   ...1

From equation (1), it follows that

v=10.2 eV×1.6×10-19 J1 eV1.6×10-27 kg×3×108 m s-1=3.4 m s-1=175 m s-1

Therefore, x=17.

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About this question

This is a previous-year question from JEE Main 2024, 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.