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) = 𝑘𝑟²/2, 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 = 𝑛ℏ, where ℏ = ℎ/(2𝜋), 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?

To determine the correct expressions, we analyze the circular motion of the particle under the influence of the central force and apply Bohr's quantization condition.

1. Equation of Motion:
For a particle of mass m moving in a circular orbit of radius r under the central force $F\left(r\right)=-kr$, the centripetal force is provided by the magnitude of the central force:
$\frac{m{v}^2}{r}=kr$
Multiplying both sides by r gives:
$m{v}^2=k{r}^2\Rightarrow v^2=\frac{k}{m}{r}^2$
Taking the square root, the speed of the particle is:
$v=r\sqrt{\frac{k}{m}}$

2. Relation for Angular Momentum:
The angular momentum L for a particle in a circular orbit of radius r with speed v is defined as:
$L=mvr$
From this definition, we can express the ratio:
$\frac{L}{m{r}^2}=\frac{mvr}{m{r}^2}=\frac{v}{r}$
Substituting $\frac{v}{r}=\sqrt{\frac{k}{m}}$ from the equation of motion, we get:
$\frac{L}{m{r}^2}=\sqrt{\frac{k}{m}}$
This confirms that the third expression is correct.

3. Bohr's Quantization Rule:
According to Bohr's quantization rule, the angular momentum is:
$L=mvr=n\hslash$
Substituting the expression for v into the quantization condition:
$m\left(r\sqrt{\frac{k}{m}}\right)r=n\hslash$
$r^2\sqrt{mk}=n\hslash$
Solving for $r^2$:
$r^2=n\hslash \sqrt{\frac{1}{mk}}$
(Note: Depending on the notation conventions, the symbol h in the options represents the reduced Planck's constant ℏ). This confirms that the first expression is correct.

4. Expression for Speed Squared:
Using the relation $v^2=\frac{k}{m}{r}^2$ and substituting the quantized value of $r^2$:
$v^2=\frac{k}{m}\left(n\hslash \sqrt{\frac{1}{mk}}\right)$
Bringing the factor $\frac{k}{m}$ inside the square root:
$v^2=n\hslash \sqrt{\frac{k^2}{m^2}\cdot \frac{1}{mk}}=n\hslash \sqrt{\frac{k}{m^3}}$
This confirms that the second expression is correct.