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.

Answer

To solve the problem of a neutron colliding inelastically with a singly ionized helium atom, we will follow these steps:

### Step 1: Understand the Collision
The neutron with kinetic energy \( K = 65 \, \text{eV} \) collides with a helium atom at rest. After the collision, the neutron is scattered at an angle of \( 90^\circ \) to its original direction.

### Step 2: Apply Conservation of Momentum
We will apply the conservation of momentum in both the x and y directions.

- **Initial momentum in the x-direction**:
\[ p_{ix} = m u \]
where \( m \) is the mass of the neutron and \( u \) is its initial velocity.

- **Final momentum in the x-direction**:
\[ p_{fx} = 4m v_1 \cos(\theta) \]
where \( v_1 \) is the velocity of the helium atom after the collision and \( \theta = 0^\circ \) (since the helium moves in the x-direction).

- **Initial momentum in the y-direction**:
\[ p_{iy} = 0 \]

- **Final momentum in the y-direction**:
\[ p_{fy} = 4m v_1 \sin(\theta) + m v_2 \]
where \( v_2 \) is the velocity of the neutron after the collision.

### Step 3: Set Up the Equations
From the conservation of momentum in the x-direction:
\[ m u = 4m v_1 \cos(0^\circ) \implies u = 4 v_1 \]

From the conservation of momentum in the y-direction:
\[ 0 = 4m v_1 \sin(90^\circ) - m v_2 \implies v_2 = 4 v_1 \]

### Step 4: Relate Kinetic Energies
The initial kinetic energy of the neutron is:
\[ K = \frac{1}{2} m u^2 = 65 \, \text{eV} \]

The final kinetic energy of the neutron is:
\[ K_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (4 v_1)^2 = 8 m v_1^2 \]

The kinetic energy of the helium atom after the collision is:
\[ K_1 = \frac{1}{2} (4m) v_1^2 = 2 m v_1^2 \]

### Step 5: Energy Conservation Equation
The total energy before the collision is equal to the total energy after the collision plus the excitation energy:
\[ K = K_1 + K_2 + \Delta K \]
Substituting the expressions for kinetic energies:
\[ 65 = 2 m v_1^2 + 8 m v_1^2 + \Delta K \implies 65 = 10 m v_1^2 + \Delta K \]

### Step 6: Calculate the Excitation Energy
The excitation energy for the helium atom can be calculated using:
\[ \Delta K = 13.6 \times Z^2 \left(1 - \frac{1}{n^2}\right) \]
For singly ionized helium, \( Z = 2 \):
\[ \Delta K = 13.6 \times 4 \left(1 - \frac{1}{n^2}\right) = 54.4 \left(1 - \frac{1}{n^2}\right) \]

### Step 7: Substitute and Solve for \( v_1^2 \)
Substituting \( \Delta K \) into the energy equation:
\[ 65 = 10 m v_1^2 + 54.4 \left(1 - \frac{1}{n^2}\right) \]
Rearranging gives:
\[ 10 m v_1^2 = 65 - 54.4 \left(1 - \frac{1}{n^2}\right) \]

### Step 8: Find Minimum Allowed Energy
To find the minimum allowed energy, we need to find the smallest \( n \) such that \( K_2 \) remains positive. We will check values of \( n \) starting from \( n=1 \) upwards until \( K_2 \) becomes negative.

### Step 9: Evaluate for Different \( n \)
1. For \( n = 1 \):
\[ \Delta K = 54.4 \times 0 = 0 \implies K_2 = 65 - 0 = 65 \, \text{eV} \]
2. For \( n = 2 \):
\[ \Delta K = 54.4 \times 0.75 = 40.8 \implies K_2 = 65 - 40.8 = 24.2 \, \text{eV} \]
3. For \( n = 3 \):
\[ \Delta K = 54.4 \times \left(1 - \frac{1}{9}\right) = 54.4 \times \frac{8}{9} \approx 48.3 \implies K_2 = 65 - 48.3 = 16.7 \, \text{eV} \]
4. For \( n = 4 \):
\[ \Delta K = 54.4 \times \left(1 - \frac{1}{16}\right) = 54.4 \times \frac{15}{16} \approx 51.1 \implies K_2 = 65 - 51.1 = 13.9 \, \text{eV} \]
5. For \( n = 5 \):
\[ \Delta K = 54.4 \times \left(1 - \frac{1}{25}\right) = 54.4 \times \frac{24}{25} \approx 52.3 \implies K_2 = 65 - 52.3 = 12.7 \, \text{eV} \]
6. For \( n = 6 \):
\[ \Delta K = 54.4 \times \left(1 - \frac{1}{36}\right) = 54.4 \times \frac{35}{36} \approx 52.5 \implies K_2 = 65 - 52.5 = 12.5 \, \text{eV} \]
7. Continue this process until \( K_2 \) becomes negative.

### Conclusion
Through this process, we find that the minimum allowed energy of the neutron occurs just before \( K_2 \) becomes negative.