In a conservative force field we can find the radial component of force from the potential energy function by using `F = -(dU)/(dr)`. Here, a positive force means repulsion and a negative force means attraction. From the given potential energy function U(r ) we can find the equilibrium position where force is zero. We can also find the ionisation energy which is the work done to move the particle from a certain position to infinity.
Let us consider a case in which a particle is bound to a certain point at a distance r from the centre of the force. The potential energy of the particle is : `U(r )=(A)/(r^(2))-(B)/(r )` where r is the distance from the centre of the force and A andB are positive constants. Answer the following questions.
The work required to move the particle from equilibrium distance to infinity is

Options

A. `(B)/(4A)`
B. `(4B)/(A)`
C. `(B^(2))/(4A)`
D. `(4B^(2))/(A)`

Answer

To solve the problem, we need to find the work required to move a particle from its equilibrium position to infinity, given the potential energy function \( U(r) = \frac{A}{r^2} - \frac{B}{r} \).

### Step-by-Step Solution:

1. **Find the Force from the Potential Energy Function**:
The force \( F \) can be derived from the potential energy function using the formula:
\[ F = -\frac{dU}{dr} \]
Given \( U(r) = \frac{A}{r^2} - \frac{B}{r} \), we differentiate \( U(r) \) with respect to \( r \).

2. **Differentiate the Potential Energy**:
\[ \frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^2}\right) - \frac{d}{dr}\left(\frac{B}{r}\right) \]
Using the power rule:
\[ \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} \]
Therefore, the force is:
\[ F = -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = \frac{2A}{r^3} - \frac{B}{r^2} \]

3. **Find the Equilibrium Position**:
At equilibrium, the net force is zero:
\[ F = 0 \implies \frac{2A}{r^3} - \frac{B}{r^2} = 0 \]
Rearranging gives:
\[ \frac{2A}{r^3} = \frac{B}{r^2} \]
Multiplying both sides by \( r^3 \):
\[ 2A = Br \implies r = \frac{2A}{B} \]

4. **Calculate the Work Done**:
The work done \( W \) to move the particle from the equilibrium position \( r_e \) to infinity is given by the change in potential energy:
\[ W = U(\infty) - U(r_e) \]
Since \( U(\infty) = 0 \), we need to calculate \( U(r_e) \):
\[ U\left(\frac{2A}{B}\right) = \frac{A}{\left(\frac{2A}{B}\right)^2} - \frac{B}{\left(\frac{2A}{B}\right)} \]
Simplifying this:
\[ U\left(\frac{2A}{B}\right) = \frac{A B^2}{4A^2} - \frac{B^2}{2A} = \frac{B^2}{4A} - \frac{B^2}{2A} \]
Combining the terms:
\[ U\left(\frac{2A}{B}\right) = \frac{B^2}{4A} - \frac{2B^2}{4A} = -\frac{B^2}{4A} \]

5. **Final Work Calculation**:
Thus, the work done is:
\[ W = 0 - \left(-\frac{B^2}{4A}\right) = \frac{B^2}{4A} \]

### Conclusion:
The work required to move the particle from the equilibrium distance to infinity is:
\[ \boxed{\frac{B^2}{4A}} \]