A particle of mass m moves in a… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Modern Physics

2019mcqhard

A particle of mass $m$ moves in a circular orbit in a central potential field $U(r)=\frac{1}{2}k{r}^{2}.$ If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number $n$ as:

Official previous-year question

Held on 12 Jan 2019 · Verified 6 Jul 2026.

Options

A
${r}_{n}\propto {n}^{2}, {E}_{n}\propto \frac{1}{{n}^{2}}$
B
${r}_{n}\propto \sqrt{n}, {E}_{n}\propto n$
C
${r}_{n}\propto n, {E}_{n}\propto n$
D
${r}_{n}\propto \sqrt{n}, {E}_{n}\propto \frac{1}{n}$

Did you get this right?

Solution

${F}_{r}=\frac{-dU}{dr}=-kr$

For circular motion

$|{F}_{r}|=kr=\frac{m{v}^{2}}{r}\Rightarrow k{r}^{2}=m{v}^{2}$ ........(i)

Bohr's quantization $\Rightarrow mvr=\frac{nh}{2\pi }$ ....(ii)

From (i) and (ii)

$\frac{{m}^{2}{v}^{2}}{m}=k{r}^{2}$

$\Rightarrow \frac{1}{m}{(\frac{nh}{2\pi r})}^{2}=k{r}^{2}\Rightarrow \frac{{n}^{2}{h}^{2}}{4{\pi }^{2}mk}={r}^{4}$

$\Rightarrow r={(\frac{{h}^{2}}{4{\pi }^{2}mk})}^{1/4} {n}^{1/2}$

$r\propto \sqrt{n}$

From equation (i) $U\propto \sqrt{n}$

$KE=\frac{1}{2}m{v}^{2}$ $PE=\frac{1}{2}k{r}^{2}$

$E=K+U=\frac{1}{2}m{v}^{2}+\frac{1}{2}k{r}^{2}=k{r}^{2}\propto n$

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Options

${r}_{n}\propto {n}^{2}, {E}_{n}\propto \frac{1}{{n}^{2}}$

${r}_{n}\propto \sqrt{n}, {E}_{n}\propto n$

${r}_{n}\propto n, {E}_{n}\propto n$

${r}_{n}\propto \sqrt{n}, {E}_{n}\propto \frac{1}{n}$

Solution

${F}_{r}=\frac{-dU}{dr}=-kr$

For circular motion

$|{F}_{r}|=kr=\frac{m{v}^{2}}{r}\Rightarrow k{r}^{2}=m{v}^{2}$ ........(i)

Bohr's quantization $\Rightarrow mvr=\frac{nh}{2\pi }$ ....(ii)

From (i) and (ii)

$\frac{{m}^{2}{v}^{2}}{m}=k{r}^{2}$

$\Rightarrow \frac{1}{m}{(\frac{nh}{2\pi r})}^{2}=k{r}^{2}\Rightarrow \frac{{n}^{2}{h}^{2}}{4{\pi }^{2}mk}={r}^{4}$

$\Rightarrow r={(\frac{{h}^{2}}{4{\pi }^{2}mk})}^{1/4} {n}^{1/2}$

$r\propto \sqrt{n}$

From equation (i) $U\propto \sqrt{n}$

$KE=\frac{1}{2}m{v}^{2}$ $PE=\frac{1}{2}k{r}^{2}$

$E=K+U=\frac{1}{2}m{v}^{2}+\frac{1}{2}k{r}^{2}=k{r}^{2}\propto n$

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Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics