Potential energy of a system of particles is `U =(alpha)/(3r^(3))-(beta)/(2r^(2)),` where r is distance between the particles. Here `alpha and beta` are positive constants. Which of the following are correct for the system?

Options

A. Equlibrium separation between the particles is `(alpha)/(beta)`
B. For `r=(alpha)/(beta),` the equlibrium is stable
C. For `r=(alpha)/(beta),` the equlibrium is unstable
D. Work required to slowly move the particles to infinite separation from initial equlibrium position is `(beta)/(alpha)`

Answer

To solve the problem, we need to analyze the potential energy function given and determine the equilibrium separation, stability of the equilibrium, and the work done in moving the particles to infinite separation.

### Step 1: Find the Equilibrium Separation
The potential energy of the system is given by:
\[ U = \frac{\alpha}{3r^3} - \frac{\beta}{2r^2} \]

To find the equilibrium separation, we need to differentiate the potential energy with respect to \( r \) and set the derivative equal to zero.

1. Differentiate \( U \) with respect to \( r \):
\[ \frac{dU}{dr} = -\frac{3\alpha}{3r^4} + \frac{\beta}{r^3} = -\frac{\alpha}{r^4} + \frac{\beta}{r^3} \]

2. Set the derivative equal to zero:
\[ -\frac{\alpha}{r^4} + \frac{\beta}{r^3} = 0 \]

3. Rearranging gives:
\[ \frac{\beta}{r^3} = \frac{\alpha}{r^4} \]

4. Cross-multiplying leads to:
\[ \beta r = \alpha \implies r = \frac{\alpha}{\beta} \]

Thus, the equilibrium separation is:
\[ r = \frac{\alpha}{\beta} \]

### Step 2: Determine Stability of the Equilibrium
To determine whether the equilibrium is stable or unstable, we need to compute the second derivative of the potential energy.

1. Differentiate \( \frac{dU}{dr} \) again:
\[ \frac{d^2U}{dr^2} = \frac{4\alpha}{3r^5} - \frac{3\beta}{r^4} \]

2. Substitute \( r = \frac{\alpha}{\beta} \):
\[ \frac{d^2U}{dr^2} = \frac{4\alpha}{3\left(\frac{\alpha}{\beta}\right)^5} - \frac{3\beta}{\left(\frac{\alpha}{\beta}\right)^4} \]

3. Simplifying:
\[ = \frac{4\alpha \beta^5}{3\alpha^5} - \frac{3\beta^5}{\alpha^4} = \frac{4\beta^5}{3\alpha^4} - \frac{3\beta^5}{\alpha^4} = \frac{\beta^5}{\alpha^4} \left(\frac{4}{3} - 3\right) = \frac{\beta^5}{\alpha^4} \left(\frac{4 - 9}{3}\right) = \frac{-5\beta^5}{3\alpha^4} \]

Since \( \frac{d^2U}{dr^2} < 0 \), the equilibrium is stable.

### Step 3: Work Done to Move to Infinite Separation
To find the work done in moving the particle to infinite separation from the equilibrium position, we calculate the potential energy at the equilibrium position and at infinity.

1. At \( r = \frac{\alpha}{\beta} \):
\[ U\left(\frac{\alpha}{\beta}\right) = \frac{\alpha}{3\left(\frac{\alpha}{\beta}\right)^3} - \frac{\beta}{2\left(\frac{\alpha}{\beta}\right)^2} \]
\[ = \frac{\alpha \beta^3}{3\alpha^3} - \frac{\beta^3}{2\alpha^2} = \frac{\beta^3}{3\alpha^2} - \frac{\beta^3}{2\alpha^2} = \frac{2\beta^3 - 3\beta^3}{6\alpha^2} = -\frac{\beta^3}{6\alpha^2} \]

2. At infinity, \( U(\infty) = 0 \).

3. The work done \( W \) is:
\[ W = U(\infty) - U\left(\frac{\alpha}{\beta}\right) = 0 - \left(-\frac{\beta^3}{6\alpha^2}\right) = \frac{\beta^3}{6\alpha^2} \]

### Conclusion
From the analysis:
- The equilibrium separation is \( \frac{\alpha}{\beta} \) (Option A is correct).
- The equilibrium is stable (Option B is incorrect).
- The work done to move to infinite separation is \( \frac{\beta^3}{6\alpha^2} \) (Option C is correct).
- Therefore, Option D is incorrect.

### Final Answers
- Correct options: A and C.