The potential energy function of a particle is given by `U(r)=A/(2r^(2))-B/(3r)`, where `A` and `B` are constant and `r` is the radial distance from the centre of the force. Choose the correct option (s)

Options

A. The equilibrium distance will be `r_(0)=(2A)/B`
B. The equilibrium distance will be `r_(0)=(3A)/B`
C. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vanish at `(r_(0))/3`, where `r_(0)` is the equilibrium distance
D. If the total energy of the particle is `(B^(2))/(6A)`, then its radial velocity will vanish at `(r_(0))/2`,where `r_(0)` is the equilibrium distance.

Explanation

To solve the problem, we need to analyze the potential energy function given by:

\[ U(r) = \frac{A}{2r^2} - \frac{B}{3r} \]

where \( A \) and \( B \) are constants, and \( r \) is the radial distance from the center of the force. We will find the equilibrium distance and check the total energy of the particle.

### Step 1: Find the Force from the Potential Energy
The force \( F \) acting on the particle can be derived from the potential energy function using the relation:

\[ F = -\frac{dU}{dr} \]

Calculating the derivative:

\[ U(r) = \frac{A}{2r^2} - \frac{B}{3r} \]

Differentiating \( U(r) \):

\[ \frac{dU}{dr} = -\frac{A}{r^3} + \frac{B}{3r^2} \]

Thus, the force is:

\[ F = -\left(-\frac{A}{r^3} + \frac{B}{3r^2}\right) = \frac{A}{r^3} - \frac{B}{3r^2} \]

### Step 2: Set the Force to Zero for Equilibrium
At equilibrium, the net force acting on the particle is zero:

\[ \frac{A}{r^3} - \frac{B}{3r^2} = 0 \]

Rearranging gives:

\[ \frac{A}{r^3} = \frac{B}{3r^2} \]

Cross-multiplying leads to:

\[ 3A = Br \]

Thus, the equilibrium distance \( r_0 \) is:

\[ r_0 = \frac{3A}{B} \]

### Step 3: Check the Total Energy at \( r = \frac{r_0}{3} \)
Given that the total energy \( E \) of the particle is \( \frac{B^2}{6A} \), we need to verify if the radial velocity vanishes at \( r = \frac{r_0}{3} \).

Substituting \( r_0 = \frac{3A}{B} \):

\[ r = \frac{r_0}{3} = \frac{3A}{3B} = \frac{A}{B} \]

### Step 4: Calculate the Potential Energy at \( r = \frac{A}{B} \)
Now, substitute \( r = \frac{A}{B} \) into the potential energy function:

\[ U\left(\frac{A}{B}\right) = \frac{A}{2\left(\frac{A}{B}\right)^2} - \frac{B}{3\left(\frac{A}{B}\right)} \]

Calculating each term:

1. First term:

\[ \frac{A}{2\left(\frac{A^2}{B^2}\right)} = \frac{AB^2}{2A^2} = \frac{B^2}{2A} \]

2. Second term:

\[ -\frac{B}{3\left(\frac{A}{B}\right)} = -\frac{B^2}{3A} \]

Combining these gives:

\[ U\left(\frac{A}{B}\right) = \frac{B^2}{2A} - \frac{B^2}{3A} = \frac{3B^2 - 2B^2}{6A} = \frac{B^2}{6A} \]

### Conclusion
The total energy \( E \) is equal to the potential energy at \( r = \frac{r_0}{3} \):

\[ E = U\left(\frac{A}{B}\right) = \frac{B^2}{6A} \]

Thus, the radial velocity vanishes at \( r = \frac{r_0}{3} \).

### Final Answers
- The equilibrium distance is \( r_0 = \frac{3A}{B} \).
- The total energy is \( \frac{B^2}{6A} \) at \( r = \frac{r_0}{3} \).