A neutron of kinetic `6.5 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 maximum allowed value of energy of the He atom?

Options

A. `13.68 eV`
B. `19.88 eV`
C. `15.26 eV`
D. `17.84 eV`

Explanation

To solve the problem step by step, we will use the conservation of energy and the concept of inelastic collisions.

### Step 1: Understand the problem
We have a neutron with a kinetic energy of \(6.5 \, \text{eV}\) colliding inelastically with a singly ionized helium atom (He\(^+\)). The neutron is scattered at an angle of \(90^\circ\) after the collision. We need to find the maximum allowed energy of the helium atom after the collision.

### Step 2: Write down the conservation of energy equation
In an inelastic collision, the total energy before the collision is equal to the total energy after the collision, including any excitation energy. The equation can be expressed as:
\[ K_n + K_{He} + E_{excitation} = K_n' + K_{He}' \]
Where:
- \(K_n\) = initial kinetic energy of the neutron = \(6.5 \, \text{eV}\)
- \(K_{He}\) = initial kinetic energy of the helium atom = \(0 \, \text{eV}\) (at rest)
- \(E_{excitation}\) = excitation energy of the helium atom
- \(K_n'\) = final kinetic energy of the neutron
- \(K_{He}'\) = final kinetic energy of the helium atom

### Step 3: Set up the equations
Since the neutron is scattered at \(90^\circ\), we can use the conservation of momentum in both the x and y directions. However, for simplicity, we will focus on the energy conservation equation.

The excitation energy for a singly ionized helium atom can be calculated using the formula:
\[ E_{excitation} = 13.6 \, \text{eV} \cdot Z^2 \left(1 - \frac{1}{n^2}\right) \]
For helium, \(Z = 2\), hence:
\[ E_{excitation} = 13.6 \cdot 4 \left(1 - \frac{1}{n^2}\right) = 54.4 \left(1 - \frac{1}{n^2}\right) \, \text{eV} \]

### Step 4: Write the energy conservation equation
The total initial energy is:
\[ K_n + K_{He} = 6.5 \, \text{eV} + 0 = 6.5 \, \text{eV} \]
The total final energy is:
\[ K_n' + K_{He}' + E_{excitation} \]
Thus, we have:
\[ 6.5 = K_n' + K_{He}' + 54.4 \left(1 - \frac{1}{n^2}\right) \]

### Step 5: Analyze the maximum energy of the helium atom
To find the maximum allowed value of the energy of the helium atom, we assume that the neutron transfers all its kinetic energy to the helium atom (which is the case for maximum energy transfer). Therefore:
\[ K_{He}' = 6.5 - E_{excitation} \]

### Step 6: Substitute the excitation energy
Substituting the excitation energy into the equation gives:
\[ K_{He}' = 6.5 - 54.4 \left(1 - \frac{1}{n^2}\right) \]

### Step 7: Solve for \(n\)
To find the maximum energy of the helium atom, we need to find the value of \(n\) that allows \(K_{He}'\) to be non-negative. This requires solving:
\[ 6.5 - 54.4 + \frac{54.4}{n^2} \geq 0 \]
This simplifies to:
\[ \frac{54.4}{n^2} \geq 47.9 \]
\[ n^2 \leq \frac{54.4}{47.9} \]
Calculating gives:
\[ n^2 \leq 1.136 \implies n \leq 1.067 \]
Since \(n\) must be a positive integer, the maximum allowed value of \(n\) is \(1\).

### Step 8: Calculate the maximum energy
Substituting \(n = 1\) back into the excitation energy equation:
\[ E_{excitation} = 54.4 \left(1 - 1\right) = 0 \, \text{eV} \]
Thus:
\[ K_{He}' = 6.5 - 0 = 6.5 \, \text{eV} \]

### Final Answer
The maximum allowed value of energy of the He atom is \(6.5 \, \text{eV}\).