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 equilibrium distance is given by

Options

A. `(A)/(B)`
B. `(2A)/(B)`
C. `(3A)/(B)`
D. `(B)/(2A)`

Answer

To find the equilibrium distance for the given potential energy function \( U(r) = \frac{A}{r^2} - \frac{B}{r} \), we will follow these steps:

### Step 1: Find the Force from the Potential Energy Function
The force \( F \) in a conservative force field is given by:
\[ F = -\frac{dU}{dr} \]
We need to differentiate the potential energy function \( U(r) \) with respect to \( r \).

### Step 2: Differentiate \( U(r) \)
Given:
\[ U(r) = \frac{A}{r^2} - \frac{B}{r} \]
We differentiate \( U(r) \):
\[ \frac{dU}{dr} = \frac{d}{dr}\left(\frac{A}{r^2}\right) - \frac{d}{dr}\left(\frac{B}{r}\right) \]
Using the power rule for differentiation:
\[ \frac{d}{dr}\left(\frac{A}{r^2}\right) = -\frac{2A}{r^3} \]
\[ \frac{d}{dr}\left(\frac{B}{r}\right) = -\frac{B}{r^2} \]
Thus,
\[ \frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2} \]

### Step 3: Substitute into the Force Equation
Now substituting into the force equation:
\[ F = -\left(-\frac{2A}{r^3} + \frac{B}{r^2}\right) = \frac{2A}{r^3} - \frac{B}{r^2} \]

### Step 4: Set the Force to Zero for Equilibrium
At equilibrium, the net force is zero:
\[ \frac{2A}{r^3} - \frac{B}{r^2} = 0 \]

### Step 5: Solve for \( r \)
Rearranging the equation:
\[ \frac{2A}{r^3} = \frac{B}{r^2} \]
Multiplying both sides by \( r^3 \):
\[ 2A = Br \]
Thus,
\[ r = \frac{2A}{B} \]

### Conclusion
The equilibrium distance \( r \) is given by:
\[ r = \frac{2A}{B} \]