Why is independent of ?

In a harmonic oscillator potential, the angular frequency is constant and independent of orbit radius, so orbital speed is strictly proportional to ( ).

JEE MainPhysics

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Atoms and Nuclei: JEE Main Physics Question with Solution

A small particle of mass

\mathrm{m}

moves in such a way that its potential energy

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

where

\omega

is constant and

\mathrm{r}

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

\mathrm{n}^{\text{th}}

orbit will be proportional to.

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What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

\sqrt{\mathrm{n}}

\mathrm{n}

\mathrm{n}^{2}

\frac{1}{n}

Step-by-step solutionView

Correct answer

The radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$ .

Option analysis

Why each option works or fails

A ·

\sqrt{\mathrm{n}}

None. This is the correct deduction. Equating the harmonic centripetal force $m\omega^2 r = mv^2/r$ gives $v = \omega r$ . Applying Bohr's condition $mvr = n\hbar$ yields $m\omega r^2 = n\hbar$ , which means $r \propto \sqrt{n}$ .

B ·

\mathrm{n}

Assuming that angular momentum being proportional to $n$ directly makes the orbital radius proportional to $n$ without accounting for the velocity's dependence on $r$ . Because the restoring force is proportional to $r$ , the speed $v$ is proportional to $r$ , making angular momentum scale with $r^2$ rather than $r$ .

C ·

\mathrm{n}^{2}

Recalling the result for the Coulomb potential (hydrogen atom), where $r_n \propto n^2$ , and applying it to a harmonic potential. Do not import hydrogen atom power laws into different potentials; derive $r(n)$ from $F = -\mathrm{d}U/\mathrm{d}r$ and $L = n\hbar$ for the specific potential given.

D ·

\frac{1}{n}

Inverting the relationship during algebraic rearrangement, mistakenly placing $n$ in the denominator. From $m\omega r^2 = n\hbar$ , solve for $r^2$ by dividing by $m\omega$ , which places $n$ in the numerator: $r^2 = \frac{n\hbar}{m\omega}$ .

Reviewed route

Solution

StepWorking

01given

Potential energy $U = \frac{1}{2}m\omega^2 r^2$ , Bohr quantization $L = mvr = \frac{nh}{2\pi}$ , circular orbit.

02find

Proportionality relation of the radius $r_n$ of the $n^{\text{th}}$ orbit with principal quantum number $n$ .

03visualise

A particle of mass $m$ executes a circular orbit of radius $r$ around the origin under a linear central restoring force provided by the harmonic oscillator potential.

04strategise

Find the restoring force using $F = -\frac{dU}{dr}$ , equate centripetal force $\frac{mv^2}{r}$ to magnitude of central force $|F|$ to find $v(r)$ , and substitute into Bohr's quantization condition $mvr = \frac{nh}{2\pi}$ .

05execute

Force magnitude $|F| = \left|-\frac{dU}{dr}\right| = m\omega^2 r$ . Centripetal balance: $\frac{mv^2}{r} = m\omega^2 r \implies v = \omega r$ . Bohr condition: $m(\omega r)r = \frac{nh}{2\pi} \implies m\omega r^2 = \frac{nh}{2\pi} \implies r = \sqrt{\frac{nh}{2\pi m\omega}} \propto \sqrt{n}$ .

✓verify

For standard Coulomb potential $U \propto -1/r$ , $r \propto n^2$ . For harmonic oscillator $U \propto r^2$ , force is proportional to $r$ , resulting in a weaker dependence $r \propto \sqrt{n}$ , which is physically consistent.

Hints that build this answer step by step

What is the central force $F(r)$ acting on the particle corresponding to the potential energy $U(r) = \frac{1}{2}m\omega^2 r^2$ ?
$F(r) = -m\omega^2 r$ , directed toward the origin
Using circular motion dynamics, how is the orbital speed $v$ related to the radius $r$ ?
$v = \omega r$
Now apply Bohr's angular momentum quantization condition $mvr = n\hbar$ with $v = \omega r$ . What is the resulting dependence of $r$ on $n$ ?
$r \propto \sqrt{n}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is

v = \omega r

independent of

n

In a harmonic oscillator potential, the angular frequency $\omega$ is constant and independent of orbit radius, so orbital speed is strictly proportional to $r$ ( $v = \omega r$ ).

Answer

The radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$ .

Why each option works or fails

A · correct: $\sqrt{\mathrm{n}}$ - None. This is the correct deduction. Equating the harmonic centripetal force $m\omega^2 r = mv^2/r$ gives $v = \omega r$ . Applying Bohr's condition $mvr = n\hbar$ yields $m\omega r^2 = n\hbar$ , which means $r \propto \sqrt{n}$ .
B: $\mathrm{n}$ - Assuming that angular momentum being proportional to $n$ directly makes the orbital radius proportional to $n$ without accounting for the velocity's dependence on $r$ . Because the restoring force is proportional to $r$ , the speed $v$ is proportional to $r$ , making angular momentum scale with $r^2$ rather than $r$ .
C: $\mathrm{n}^{2}$ - Recalling the result for the Coulomb potential (hydrogen atom), where $r_n \propto n^2$ , and applying it to a harmonic potential. Do not import hydrogen atom power laws into different potentials; derive $r(n)$ from $F = -\mathrm{d}U/\mathrm{d}r$ and $L = n\hbar$ for the specific potential given.
D: $\frac{1}{n}$ - Inverting the relationship during algebraic rearrangement, mistakenly placing $n$ in the denominator. From $m\omega r^2 = n\hbar$ , solve for $r^2$ by dividing by $m\omega$ , which places $n$ in the numerator: $r^2 = \frac{n\hbar}{m\omega}$ .

Step-by-step solution

given: Potential energy $U = \frac{1}{2}m\omega^2 r^2$ , Bohr quantization $L = mvr = \frac{nh}{2\pi}$ , circular orbit.
find: Proportionality relation of the radius $r_n$ of the $n^{\text{th}}$ orbit with principal quantum number $n$ .
visualise: A particle of mass $m$ executes a circular orbit of radius $r$ around the origin under a linear central restoring force provided by the harmonic oscillator potential.
strategise: Find the restoring force using $F = -\frac{dU}{dr}$ , equate centripetal force $\frac{mv^2}{r}$ to magnitude of central force $|F|$ to find $v(r)$ , and substitute into Bohr's quantization condition $mvr = \frac{nh}{2\pi}$ .
execute: Force magnitude $|F| = \left|-\frac{dU}{dr}\right| = m\omega^2 r$ . Centripetal balance: $\frac{mv^2}{r} = m\omega^2 r \implies v = \omega r$ . Bohr condition: $m(\omega r)r = \frac{nh}{2\pi} \implies m\omega r^2 = \frac{nh}{2\pi} \implies r = \sqrt{\frac{nh}{2\pi m\omega}} \propto \sqrt{n}$ .
verify: For standard Coulomb potential $U \propto -1/r$ , $r \propto n^2$ . For harmonic oscillator $U \propto r^2$ , force is proportional to $r$ , resulting in a weaker dependence $r \propto \sqrt{n}$ , which is physically consistent.

Atoms and Nuclei: JEE Main Physics Question with Solution

What feels right?

The radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$ .

Why each option works or fails

Solution

01given

02find

03visualise

04strategise

05execute

✓verify

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