JEE Main 2019PhysicsAtomic PhysicsHardMCQ

JEE Main 2019 — Atomic Physics Question with Solution

JEE Main 2019 (12 Jan Shift 1)

Question

A particle of mass m moves in a circular orbit in a central potential field Ur=12kr2. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

Choose an option

Show full solutionCorrect option: B
Correct answer
Brn∝n, En∝n

Step-by-step explanation

Fr=-dUdr=-kr

For circular motion

|Fr|=kr=mv2r⇒kr2=mv2 ........(i)

Bohr's quantization ⇒mvr=nh2π ....(ii)

From (i) and (ii)

m2v2m=kr2

⇒1mnh2πr2=kr2⇒n2h24π2mk=r4

⇒r=h24π2mk1/4 n1/2

r∝n

From equation (i) U∝n

KE=12mv2; PE=12kr2

E=K+U=12mv2+12kr2=kr2∝n

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Atomic Physics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.