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 , 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 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?(A) r 2 = n h sqrt (1 /m k), (B) v 2 = n h √ k /m 3 (C) L /m r 2 = √ k/ m(D) E = n h /2 (√ k /m)

 Alright, let's tackle this problem step by step. 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 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.

Understanding the Central Force and Circular Motion

First, let's recall that for a particle moving in a circular orbit under a central force, the centripetal force required to keep the particle in circular motion is provided by the central force itself.

The centripetal force is given by:

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

The given central force is:

$$F(r)=-kr$$

Since the force is attractive (negative sign indicates direction towards the center), the magnitude provides the centripetal force:

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

Angular Momentum Quantization

Bohr's quantization rule states:

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

where $L$ is the angular momentum, $\mathrm{\hslash }=\frac{h}{2\pi }$, and $n$ is a positive integer.

The angular momentum for a particle in circular motion is:

$$L=mvr$$

So,

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

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 $mr$:

$$m{r}^2\sqrt{\frac{k}{m}}=n\mathrm{\hslash }$$$$r^2\sqrt{mk}=n\mathrm{\hslash }$$$$r^2=\frac{n\mathrm{\hslash }}{\sqrt{mk}}$$$$r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}\text{(3)}$$

This looks like option (A), but let's check the exact form:
Option (A): $r^2=n\mathrm{\hslash }\sqrt{\frac{1}{mk}}$

Yes, this matches equation (3). So, option (A) is correct.

Finding $v^2$

From equation (1):

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

Square both sides:

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

From equation (3):

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

So,

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

Now, compare with option (B):
Option (B): $v^2=n\mathrm{\hslash }\sqrt{\frac{k}{m^3}}$

This matches exactly. So, option (B) is correct.

Checking $\frac{L}{m{r}^2}$

Given:

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

From equation (3):

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

So,

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

Now, compare with option (C):
Option (C): $\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$

This matches exactly. So, option (C) is correct.

Total Energy $E$

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

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

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 equation (3):

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

So,

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

Now, compare with option (D):
Option (D): $E=\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$

This is half of what we derived. Our calculation shows $E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$, not $\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$.

Wait, let me re-examine the energy calculation.

Actually, for a harmonic oscillator (which this is, since $F=-kr$), the total energy is equally divided between kinetic and potential energy. So, $E=2\times \frac{1}{2}k{r}^2=k{r}^2$, as I had. But the standard quantum harmonic oscillator has energy levels $E_n=\mathrm{\hslash }\omega (n+\frac{1}{2})$, where $\omega =\sqrt{\frac{k}{m}}$.

But here, Bohr's quantization is applied differently, leading to $E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$. So, option (D) is missing a factor of 2 compared to our result. Thus, option (D) is incorrect as stated.

Re-evaluating Option (D)

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

From our earlier steps, $E=n\mathrm{\hslash }\sqrt{\frac{k}{m}}$, whereas option (D) has $\frac{n\mathrm{\hslash }}{2}\sqrt{\frac{k}{m}}$. So, unless there's a factor I'm missing, option (D) seems incorrect.

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 (should be $n\mathrm{\hslash }\sqrt{\frac{k}{m}}$)

Final Answer

The correct options are (A), (B), and (C).

Expressed in the required format:

• Correct options: $\boxed{{\displaystyle A,B,C}}$