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.
If the total energy of the particle is `E=-(3B^(2))/(16A)`, and it is known that the motion is radial only then the velocity is zero at

Options

A. (a)`(r_(0))/(3)`
B. (b)`(2r_(0))/(3)`
C. (c)`r_(0)`
D. (d)`(2r_(0))/(5)`

Answer

To solve the problem, we need to find the position at which the velocity of the particle is zero, given the potential energy function and the total energy of the particle. Let's go through the solution step by step.

### Step 1: Write down the potential energy function and total energy
The potential energy function is given as:
\[ U(r) = \frac{A}{r^2} - \frac{B}{r} \]
The total energy \( E \) of the particle is given as:
\[ E = -\frac{3B^2}{16A} \]

### Step 2: Set up the equation for total energy
In a conservative force field, the total energy \( E \) is the sum of kinetic energy \( K \) and potential energy \( U \):
\[ E = K + U \]
Since the motion is radial only, the kinetic energy can be expressed as:
\[ K = \frac{1}{2} m v^2 \]
At the point where the velocity is zero, \( v = 0 \), thus \( K = 0 \). Therefore, we have:
\[ E = U(r) \]

### Step 3: Equate total energy to potential energy
Substituting the expression for total energy:
\[ -\frac{3B^2}{16A} = \frac{A}{r^2} - \frac{B}{r} \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ \frac{A}{r^2} - \frac{B}{r} + \frac{3B^2}{16A} = 0 \]

### Step 5: Multiply through by \( 16Ar^2 \) to eliminate denominators
Multiplying through by \( 16Ar^2 \) results in:
\[ 16A^2 - 16ABr + 3B^2r^2 = 0 \]

### Step 6: Rearrange into standard quadratic form
Rearranging gives us a standard quadratic equation:
\[ 3B^2r^2 - 16ABr + 16A^2 = 0 \]

### Step 7: Apply the quadratic formula
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3B^2 \), \( b = -16AB \), and \( c = 16A^2 \):
\[ r = \frac{16AB \pm \sqrt{(-16AB)^2 - 4 \cdot 3B^2 \cdot 16A^2}}{2 \cdot 3B^2} \]
Calculating the discriminant:
\[ (-16AB)^2 - 4 \cdot 3B^2 \cdot 16A^2 = 256A^2B^2 - 192A^2B^2 = 64A^2B^2 \]
Thus, we have:
\[ r = \frac{16AB \pm 8AB}{6B^2} \]

### Step 8: Solve for the two possible values of r
This gives us two solutions:
1. \( r = \frac{24AB}{6B^2} = \frac{4A}{B} \)
2. \( r = \frac{8AB}{6B^2} = \frac{4A}{3B} \)

### Step 9: Identify the equilibrium position
Since we are interested in the position where the velocity is zero, we take the physically meaningful solution:
\[ r = \frac{4A}{3B} \]

### Final Answer:
Thus, the position where the velocity is zero is:
\[ r = \frac{4A}{3B} \]