A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Befor 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?

Explanation

To solve the problem, we need to determine the minimum kinetic energy of a moving hydrogen atom such that, after a head-on collision with a stationary hydrogen atom, one of the atoms reaches an excited state.

### Step-by-Step Solution:

1. **Understanding the Energy Levels**:
The energy difference between the ground state (n=1) and the first excited state (n=2) of a hydrogen atom is given by:
\[ \Delta E = E_2 - E_1 = 10.2 \text{ eV} \]

2. **Conservation of Momentum**:
Before the collision, we have one moving hydrogen atom with kinetic energy \( K \) and one stationary hydrogen atom. After the collision, both atoms move together. By conservation of momentum:
\[ m v = (2m) v' \]
where \( v \) is the velocity of the moving atom before the collision and \( v' \) is the velocity of both atoms after the collision. This gives us:
\[ v' = \frac{v}{2} \]

3. **Relating Kinetic Energies**:
The initial kinetic energy \( K \) of the moving atom can be expressed in terms of the velocity \( v \):
\[ K = \frac{1}{2} m v^2 \]
After the collision, the kinetic energy of both atoms moving together is:
\[ K' = \frac{1}{2} (2m) v'^2 = \frac{1}{2} (2m) \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \frac{v^2}{2} = \frac{K}{2} \]

4. **Conservation of Energy**:
The total energy before the collision must equal the total energy after the collision. Therefore:
\[ K = K' + \Delta E \]
Substituting \( K' = \frac{K}{2} \):
\[ K = \frac{K}{2} + \Delta E \]

5. **Solving for K**:
Rearranging the equation gives:
\[ K - \frac{K}{2} = \Delta E \]
\[ \frac{K}{2} = \Delta E \]
\[ K = 2 \Delta E \]
Substituting \( \Delta E = 10.2 \text{ eV} \):
\[ K = 2 \times 10.2 \text{ eV} = 20.4 \text{ eV} \]

### Final Answer:
The minimum value of the kinetic energy \( K \) of the moving hydrogen atom is:
\[ \boxed{20.4 \text{ eV}} \]