JEE Main 2023 · PhysicsMultiple choiceSingle correctMediumMulti-step

JEE Main 6 April 2023, Shift 2, Physics Q43

Question 43 of 90 in this shift, Physics question 13 of 30, Section A.

A small particle of mass

m

moves in such a way that its potential energy

U=\frac{1}{2}m\omega^2 r^2

where

\omega

is constant and

r

is the distance of the particle from origin. Assuming Bohr's quantization of momentum and circular orbit, the radius of

n^{\mathrm{th}}

orbit will be proportional to,

(1) $n$
(2) $n^2$
(3) $\sqrt{n}$ Official answer
(4) $\frac{1}{n}$

Official answer

Option 3

NTA final key.

Chapter: Atoms
Topic: Bohr's Atomic Model
Idea tested: Angular Momentum Quantization

Same idea in other shifts

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.