JEE Main 2019PhysicsAtomic PhysicsEasyMCQ

JEE Main 2019 — Atomic Physics Question with Solution

JEE Main 2019 (12 Apr Shift 1)

Question

An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm in making a transition to the ground state. The quantum number n, corresponding to its initial excited state is

(for a photon of wavelength λ, energy E=1240 eVλ(in nm))

Choose an option

Show full solutionCorrect option: B
Correct answer
Bn=5

Step-by-step explanation

E1=1240λ1=1240108.5≈11.43 eV
E2=1240λ2=124030.4≈40.79 eV
ETotal=E1+E1=52.22 eV
52.22=13.6221-1n2
52.22=54.41-1n2
0.96=1-1n2
1n2=0.04
n2=10.04=1004=25
n=5

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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.