JEE Main 2021PhysicsAtomic PhysicsHardMCQ

JEE Main 2021 — Atomic Physics Question with Solution

JEE Main 2021 (20 Jul Shift 1)

Question

The radiation corresponding to 3→2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of 5×10-4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is:
(Mass of electron =9.1×10-31 kg)

Choose an option

Show full solutionCorrect option: D
Correct answer
D0.82 eV

Step-by-step explanation

3→2⇒1.89 eV
5×10-4 T  r=7 mm
r=mvqB⇒mv=qrB
⇒E=P22 m=(qRB)22 m
=1.6×10-19×7×10-3×5×10-422×9.1×10-31Joule
=3136×10-5218.2×10-31×1.6×10-19 eV
=1.077 eV
We know work function = energy incident - (KE)electron 

ϕ=1.89-1.077=0.813 eV

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About this question

This is a previous-year question from JEE Main 2021, 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.