A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excited state?

Options

A. `20.4eV`
B. `10.2eV`
C. `54.4eV`
D. `13.6eV`

Explanation

To solve the problem of determining the minimum kinetic energy of a moving hydrogen atom such that one of the atoms reaches an excited state after a head-on collision with a stationary hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the System**: We have two hydrogen atoms, one moving with kinetic energy \( K \) and the other stationary. Both are initially in the ground state.

2. **Conservation of Momentum**:
- Before the collision, the momentum of the system is given by the moving atom:
\[ p_{\text{initial}} = mv \]
- After the collision, both atoms move together with a common velocity \( v' \). The total momentum after the collision is:
\[ p_{\text{final}} = 2m v' \]
- By conservation of momentum, we have:
\[ mv = 2m v' \implies v' = \frac{v}{2} \]

3. **Calculate Kinetic Energy After Collision**:
- The initial kinetic energy \( K \) of the moving atom is:
\[ K = \frac{1}{2} mv^2 \]
- The kinetic energy after the collision, when both atoms move together, is:
\[ K' = \frac{1}{2} (2m) v'^2 = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2 = \frac{1}{2} (2m) \frac{v^2}{4} = \frac{mv^2}{4} \]

4. **Determine the Energy Loss**:
- The energy lost in the collision is:
\[ \Delta K = K - K' = \frac{1}{2} mv^2 - \frac{mv^2}{4} = \frac{1}{4} mv^2 \]

5. **Energy Required for Excitation**:
- The minimum energy required to excite a hydrogen atom from the ground state to the first excited state is 10.2 eV.

6. **Set Up the Equation**:
- The energy lost during the collision must be equal to or greater than the energy required for excitation:
\[ \frac{1}{4} mv^2 \geq 10.2 \text{ eV} \]

7. **Solve for Kinetic Energy \( K \)**:
- Rearranging gives:
\[ mv^2 \geq 40.8 \text{ eV} \]
- Since \( K = \frac{1}{2} mv^2 \), we have:
\[ K \geq \frac{40.8}{2} \text{ eV} = 20.4 \text{ eV} \]

### Final Answer:
The minimum value of the kinetic energy of the moving hydrogen atom such that one of the atoms reaches an excited state is **20.4 eV**.