A neutron moving with a speed `v` makes a head-on collision with a hydrogen in ground state kept at rest which inelastic collision will be take place is (assume that mass of photon is nearly equal to the mass of neutron)

Options

A. `10.2 eV`
B. `20.4 eV`
C. `12.1 eV`
D. `16.8 eV`

Answer

To solve the problem of a neutron colliding with a hydrogen atom in an inelastic collision, we will follow these steps:

### Step 1: Understand the problem
We have a neutron moving with a speed \( v \) that collides head-on with a hydrogen atom at rest. The collision is inelastic, meaning that some kinetic energy will be converted into other forms of energy (like internal energy of the hydrogen atom).

### Step 2: Apply conservation of momentum
The conservation of momentum states that the total momentum before the collision must equal the total momentum after the collision.

Before the collision:
- Momentum of neutron = \( mv \) (where \( m \) is the mass of the neutron)
- Momentum of hydrogen = \( 0 \) (since it is at rest)

After the collision:
- Let the speed of the neutron after the collision be \( v_1 \) and the speed of the hydrogen atom be \( v_2 \).

The equation for conservation of momentum can be written as:
\[ mv = mv_1 + mv_2 \]
Dividing through by \( m \) gives:
\[ v = v_1 + v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply conservation of kinetic energy
In an inelastic collision, some kinetic energy is lost. The kinetic energy before the collision is:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]
The kinetic energy after the collision is:
\[ KE_{\text{final}} = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2 \]
Let \( \Delta E \) be the energy lost in the collision. Thus, we can write:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2 + \Delta E \]
Dividing through by \( \frac{1}{2} m \) gives:
\[ v^2 = v_1^2 + v_2^2 + \frac{2\Delta E}{m} \quad \text{(Equation 2)} \]

### Step 4: Relate the equations
From Equation 1, we can square both sides:
\[ v^2 = (v_1 + v_2)^2 = v_1^2 + v_2^2 + 2v_1v_2 \quad \text{(Equation 3)} \]

### Step 5: Set Equations 2 and 3 equal
Since both equations equal \( v^2 \), we can set them equal to each other:
\[ v_1^2 + v_2^2 + \frac{2\Delta E}{m} = v_1^2 + v_2^2 + 2v_1v_2 \]
This simplifies to:
\[ \frac{2\Delta E}{m} = 2v_1v_2 \]
Dividing both sides by 2 gives:
\[ \frac{\Delta E}{m} = v_1v_2 \quad \text{(Equation 4)} \]

### Step 6: Ensure real speeds
For \( v_1 \) and \( v_2 \) to be real numbers, the expression \( v_1 - v_2 \) must be non-negative:
\[ (v_1 - v_2)^2 \geq 0 \]
Expanding this gives:
\[ v_1^2 - 2v_1v_2 + v_2^2 \geq 0 \]
Substituting \( v_1^2 + v_2^2 \) from Equation 3, we find:
\[ v^2 - 4v_1v_2 \geq 0 \]
This implies:
\[ v^2 \geq 4v_1v_2 \]
Substituting Equation 4 gives:
\[ v^2 \geq 4 \frac{\Delta E}{m} \]
Multiplying through by \( m \):
\[ mv^2 \geq 4\Delta E \]

### Step 7: Calculate minimum kinetic energy
For hydrogen, the minimum energy required to excite it from the ground state to the first excited state is \( \Delta E = 10.2 \) eV. Thus:
\[ mv^2 \geq 4 \times 10.2 \text{ eV} = 40.8 \text{ eV} \]
Therefore, the minimum kinetic energy required is:
\[ \frac{1}{2} mv^2 \geq 20.4 \text{ eV} \]

### Final Answer
The minimum kinetic energy required for the neutron to cause an inelastic collision with the hydrogen atom is \( 20.4 \) eV.

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