Alright, let's tackle this problem step by step. I'm going to break it down to understand which of the given expressions for the particle's radius 

$r$, speed $v$, angular momentum ratio $\frac{L}{m{r}^2}$, and total energy $E$ are correct under the given conditions.

Understanding the Problem

We have a particle of mass $m$ moving in a circular orbit under a central force $F(r)=-kr$, which corresponds to the potential energy $V(r)=\frac{1}{2}k{r}^2$. The angular momentum $L$ of the particle is quantized according to Bohr's quantization rule:

$$L=n\mathrm{\hslash }$$

where $\mathrm{\hslash }=\frac{h}{2\pi }$, $h$ is Planck's constant, and $n$ is a positive integer.

We need to determine which of the given expressions for $r$, $v$, $\frac{L}{m{r}^2}$, and $E$ are correct.

Given Options

Let's list the options we need to evaluate:

(A) $r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$

(B) $v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$

(C) $\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$

(D) $E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$

Step 1: Circular Motion Under Central Force

For a particle moving in a circular orbit under a central force $F(r)=-kr$, the centripetal force required to keep the particle in circular motion is provided by this central force.

The centripetal force is:

$$F_{\text{centripetal}}=\frac{m{v}^2}{r}$$

Given the central force:

$$F(r)=-kr$$

The magnitude of the force is $kr$, so:

$$kr=\frac{m{v}^2}{r}$$$$k{r}^2=m{v}^2$$$$v^2=\frac{k{r}^2}{m}$$$$v=r\sqrt{\frac{k}{m}}\text{(1)}$$

Step 2: Angular Momentum Quantization

Bohr's quantization rule states:

$$L=n\mathrm{\hslash }$$

The angular momentum $L$ for a particle in circular motion is:

$$L=mvr$$

So:

$$mvr=n\mathrm{\hslash }$$$$v=\frac{n\mathrm{\hslash }}{mr}\text{(2)}$$

Step 3: Relating Equations (1) and (2)

From equation (1):

$$v=r\sqrt{\frac{k}{m}}$$

From equation (2):

$$v=\frac{n\mathrm{\hslash }}{mr}$$

Set them equal:

$$r\sqrt{\frac{k}{m}}=\frac{n\mathrm{\hslash }}{mr}$$

Multiply both sides by $mr$:

$$m{r}^2\sqrt{\frac{k}{m}}=n\mathrm{\hslash }$$$$r^2\sqrt{mk}=n\mathrm{\hslash }$$$$r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$$$$r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$$

This matches option (A):

$$\text{(A)}{r}^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}\text{is correct.}$$

Step 4: Finding $v^2$

From equation (1):

$$v=r\sqrt{\frac{k}{m}}$$

Square both sides:

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

From the expression for $r^2$ we just found:

$$r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$$

So:

$$v^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}\cdot \frac{k}{m}$$$$v^2=n\mathrm{\hslash }\cdot \frac{k}{m\sqrt{mk}}$$$$v^2=n\mathrm{\hslash }\cdot \frac{\sqrt{k}}{\sqrt{m}\cdot m}$$$$v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$$

This matches option (B):

$$\text{(B)}{v}^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}\text{is correct.}$$

Step 5: Checking $\frac{L}{m{r}^2}$

Given:

$$L=n\mathrm{\hslash }$$

From the expression for $r^2$:

$$r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$$

So:

$$\frac{L}{m{r}^2}=\frac{n\mathrm{\hslash }}{m\cdot \frac{n\mathrm{\hslash }}{\sqrt{mk}}}$$$$\frac{L}{m{r}^2}=\frac{n\mathrm{\hslash }\cdot \sqrt{mk}}{mn\mathrm{\hslash }}$$$$\frac{L}{m{r}^2}=\frac{\sqrt{mk}}{m}$$$$\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$$

This matches option (C):

$$\text{(C)}\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}\text{is correct.}$$

Step 6: Total Energy $E$

The total energy $E$ is the sum of kinetic energy $T$ and potential energy $V$:

$$E=T+V$$$$T=\frac{1}{2}m{v}^2$$

From equation (1):

$$v=r\sqrt{\frac{k}{m}}$$

So:

$$v^2=r^2\frac{k}{m}$$$$T=\frac{1}{2}m\cdot r^2\frac{k}{m}=\frac{1}{2}k{r}^2$$

Potential energy:

$$V=\frac{1}{2}k{r}^2$$

So:

$$E=\frac{1}{2}k{r}^2+\frac{1}{2}k{r}^2=k{r}^2$$

From the expression for $r^2$:

$$r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$$

So:

$$E=k\cdot \frac{n\mathrm{\hslash }}{\sqrt{mk}}$$$$E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$$

Now, compare with option (D):

$$\text{(D)}E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$$

Our calculation shows $E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$, not $\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$. Therefore, option (D) is incorrect.

Step 7: Verifying Energy Calculation

Wait a minute, let me double-check the energy calculation because for a harmonic oscillator, the total energy is equally divided between kinetic and potential energy, so $E=2\times \frac{1}{2}k{r}^2=k{r}^2$, which seems correct.

Alternatively, in quantum mechanics, the energy levels of a harmonic oscillator are $E_n=\mathrm{\hslash }\omega (n+\frac{1}{2})$, where $\omega =\sqrt{\frac{k}{m}}$. But here, we're using Bohr's quantization, which might differ.

Given that, and our consistent derivation leading to $E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$, option (D) seems to be missing a factor of 2 compared to our result. Thus, option (D) is incorrect as stated.

Conclusion

After carefully analyzing each option:

• (A) $r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$ → Correct

• (B) $v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$ → Correct

• (C) $\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$ → Correct

• (D) $E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$ → Incorrect

Final Answer

The correct options are A, B, and C.

• A particle of mass m is moving in a circular orbit under the influence of the central force F(r) = -kr, corresponding to the potential energy V ( r ) = k r 2 / 2 𝑉 ( 𝑟 ) = 𝑘 𝑟 2 / 2 , where k is a positive force constant and r is the radial distance from the origin.  According to the Bohr's quantization rule, the angular momentum of the particle is given by L = nh, where   h = h / ( 2 π ) , h ℎ = ℎ / ( 2 𝜋 ) , ℎ is the Planck's constant, and n a positive integer. If v and E are the speed and total energy of the  particle, respectively, then which of the following expression(s) is(are) correct