1. Given data

A particle moves in a circular orbit under the central force $F(r)=-kr$ with potential energy $V(r)=\frac{1}{2}kr^2.$

Also, Bohr quantization gives $L=n\hbar.$

We must check each option.

2. Condition for circular motion

For a circular orbit of radius $r$, the required centripetal force is $\frac{mv^2}{r}.$

Since the given central force provides this, $\frac{mv^2}{r}=kr.$ So, $mv^2=kr^2$ v^2=\frac{k}{m}r^2. \tag{1}

3. Use angular momentum quantization

For circular motion, L=mvr=n\hbar. \tag{2}

Using (1), $v=r\sqrt{\frac{k}{m}}.$

Substitute into (2): $L=m\left(r\sqrt{\frac{k}{m}}\right)r$ $L=r^2\sqrt{mk}.$

Now using $L=n\hbar$, $r^2\sqrt{mk}=n\hbar$ $r^2=n\hbar\frac{1}{\sqrt{mk}}=n\hbar\sqrt{\frac{1}{mk}}.$

So Option A is correct.

4. Check option B

From (1), $v^2=\frac{k}{m}r^2.$ Substitute the value of $r^2$ from option A result: $v^2=\frac{k}{m}\cdot n\hbar\sqrt{\frac{1}{mk}}.$

Simplify: $v^2=n\hbar\sqrt{\frac{k}{m^3}}.$

So Option B is correct.

5. Check option C

We found above that $L=r^2\sqrt{mk}.$ Hence, $\frac{L}{mr^2}=\frac{r^2\sqrt{mk}}{mr^2}=\sqrt{\frac{k}{m}}.$

So Option C is correct.

6. Check option D

Total energy is $E=T+V.$

Kinetic energy: $T=\frac{1}{2}mv^2.$ Using $mv^2=kr^2$, $T=\frac{1}{2}kr^2.$

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

Therefore, $E=kr^2.$

Now substitute $r^2=n\hbar\sqrt{\frac{1}{mk}},$ so $E=k\cdot n\hbar\sqrt{\frac{1}{mk}}=n\hbar\sqrt{\frac{k}{m}}.$

But option D says $E=\frac{n\hbar}{2}\sqrt{\frac{k}{m}},$ which is smaller by a factor of $2$.

So Option D is incorrect.

7. Final answer

The correct expressions are:

• A
• B
• C