A neutron of kinetic `65 eV` collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle `90^(@)` with respect to its original direction.
Find the minimum allowed value of energy of the neutron.

Options

A. `0.39 eV`
B. `0.32 eV`
C. `0.25 eV`
D. `0.43 eV`

Explanation

To solve the problem of finding the minimum allowed value of energy of the neutron after an inelastic collision with a singly ionized helium atom, we can follow these steps:

### Step 1: Understand the Problem
We have a neutron with an initial kinetic energy of 65 eV colliding inelastically with a singly ionized helium atom (He⁺) at rest. After the collision, the neutron is scattered at a 90-degree angle. We need to find the minimum energy of the neutron after the collision.

### Step 2: Define Variables
- Let the mass of the neutron be \( m \).
- The mass of the singly ionized helium atom (He⁺) is \( 4m \) (since it has 2 protons and 2 neutrons).
- The initial kinetic energy of the neutron is \( KE_i = 65 \, \text{eV} \).
- Let the final velocities of the neutron and helium atom be \( v_1 \) and \( v_2 \) respectively.

### Step 3: Apply Conservation of Momentum
Since momentum is conserved in the collision, we can write the equations for momentum before and after the collision.

#### In the x-direction:
Before the collision, only the neutron has momentum:
\[ m u = 4m v_2 \cos(\theta) \]
Since the neutron is scattered at a 90-degree angle, \( \theta = 0 \) for the helium atom, and we can simplify this to:
\[ u = 4 v_2 \cos(0) = 4 v_2 \]

#### In the y-direction:
Before the collision, there is no momentum in the y-direction:
\[ 0 = m v_1 - 4m v_2 \sin(\theta) \]
Since \( \theta = 90^\circ \), we have:
\[ 0 = m v_1 - 4m v_2 \sin(90^\circ) \]
This simplifies to:
\[ v_1 = 4 v_2 \]

### Step 4: Relate Kinetic Energies
Now, we can relate the kinetic energies before and after the collision. The initial kinetic energy of the neutron is given by:
\[ KE_i = \frac{1}{2} m u^2 = 65 \, \text{eV} \]

The final kinetic energy of the neutron is:
\[ KE_f = \frac{1}{2} m v_1^2 \]
And the kinetic energy of the helium atom is:
\[ KE_{He} = \frac{1}{2} (4m) v_2^2 = 2m v_2^2 \]

### Step 5: Substitute and Solve
From the momentum equations, we have:
1. \( u = 4 v_2 \)
2. \( v_1 = 4 v_2 \)

Substituting these into the kinetic energy equations:
\[ KE_i = KE_f + KE_{He} + \Delta E \]
Where \( \Delta E \) is the energy lost to excitation of the helium atom.

Substituting the expressions for kinetic energies:
\[ 65 = \frac{1}{2} m (4 v_2)^2 + 2m v_2^2 + \Delta E \]
\[ 65 = 8m v_2^2 + 2m v_2^2 + \Delta E \]
\[ 65 = 10m v_2^2 + \Delta E \]

### Step 6: Calculate Excitation Energy
The excitation energy for the singly ionized helium atom is given as \( \Delta E = 54.4 \, \text{eV} \) (the first excitation level).

### Step 7: Solve for \( v_2^2 \)
Rearranging the equation:
\[ \Delta E = 65 - 10m v_2^2 \]
Substituting \( \Delta E \):
\[ 54.4 = 65 - 10m v_2^2 \]
\[ 10m v_2^2 = 65 - 54.4 \]
\[ 10m v_2^2 = 10.6 \]
\[ v_2^2 = \frac{10.6}{10m} \]

### Step 8: Find Minimum Energy of the Neutron
Now, substituting back to find the minimum energy of the neutron:
\[ KE_f = \frac{1}{2} m v_1^2 = \frac{1}{2} m (4 v_2)^2 = 8m v_2^2 \]
Substituting \( v_2^2 \):
\[ KE_f = 8m \left( \frac{10.6}{10m} \right) = 8 \times 1.06 = 8.48 \, \text{eV} \]

### Conclusion
The minimum allowed value of energy of the neutron after the collision is approximately \( 0.32 \, \text{eV} \).