Given potential

The central potential is $U(r)=\frac{1}{2}kr^2$

For a circular orbit in a central force field, the required centripetal force is provided by the central force.

Find the force

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

Its magnitude is $|F|=kr$

Condition for circular motion

For a particle moving in a circle of radius $r$ with speed $v$, $\frac{mv^2}{r}=kr$

So, $mv^2=kr^2$ $v^2=\frac{k}{m}r^2$ $v=r\sqrt{\frac{k}{m}}$

Apply Bohr quantization

Bohr's quantization condition is $mvr=n\hbar$

Substitute $v=r\sqrt{k/m}$: $m\left(r\sqrt{\frac{k}{m}}\right)r=n\hbar$

$\sqrt{mk}\,r^2=n\hbar$

Hence, $r^2\propto n$ $r_n\propto \sqrt{n}$

Find total energy

Total energy is $E=T+U$

Kinetic energy: $T=\frac{1}{2}mv^2$

Using $mv^2=kr^2$, $T=\frac{1}{2}kr^2$

Potential energy: $U=\frac{1}{2}kr^2$

Therefore, $E=\frac{1}{2}kr^2+\frac{1}{2}kr^2=kr^2$

Since $r^2\propto n$, $E_n\propto n$

Match with options

We found: $r_n\propto \sqrt{n}, \qquad E_n\propto n$

This matches Option A.

Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.