Verified Solution • 2025–26 Session moderate

QJEEAdv24Ph68ca: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph68ca Solution: A particle of mass is moving in a circular orbit under the…

A particle of mass mm is moving in a circular orbit under the influence of the central force F(r)=−krF(r)=-k r , corresponding to the potential energy V(r)=kr2/2V(r)=k r^2 / 2 , where kk is a positive force constant and rr is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=nℏL=n \hbar , where ℏ=h/(2π),h\hbar=h /(2 \pi), h is the Planck's constant, and nn a positive integer. If vv and EE are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Multiple Choice Options

  • A r2=nℏ1mkr^2=n \hbar \sqrt{\frac{1}{m k}}
  • B v2=nℏkm3v^2=n \hbar \sqrt{\frac{k}{m^3}}
  • C Lmr2=km\frac{L}{m r^2}=\sqrt{\frac{k}{m}}
  • D E=nℏ2kmE=\frac{n \hbar}{2} \sqrt{\frac{k}{m}}

Step-by-Step Verified Solution

The central force will provide necessary centripetal force

⇒kr=mv2r\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}

or, kr2=mv2... (1)\mathrm{kr}^2=\mathrm{mv}^2\quad \text{... (1)}

By quantisation rule nℏ=mvr\mathrm{n} \hbar=\mathrm{mvr}

or, nℏr=mv... (2)\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}\quad \text{... (2)}

amp;(1)(2)2⇒kr2n2ℏ2r2=mv2 m2v2amp;⇒kn2ℏ2r4=1 mamp;⇒r=(n2ℏ2 km)14⇒r2=nℏmk\begin{aligned}& \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\& \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\& \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}\end{aligned}

(B) Using (1), K⋅nℏmk=mv2\mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2

⇒v2=nℏkm3\Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}

(C) Lmr2=mvrmr2=vr=km\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} from (1)

(D) E=12mv2+12kr2=nℏ2km+12knℏmk\mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}

E=nℏkm\mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring