An electron of stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquires as a result of photon emission is `k h R`, Find `k .` (Here, `m=` Mass of electron, `h=` Planck's `overline(25 m)` constant and `R=` Rydberg constant )

Answer

To solve the problem, we need to find the value of \( k \) when an electron transitions from the fifth energy level (\( n=5 \)) to the ground state (\( n=1 \)) in a hydrogen atom, and the velocity acquired by the atom due to photon emission is given as \( k h R \).

### Step-by-Step Solution:

1. **Identify the Energy Levels**:
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{Z^2 \cdot R_H}{n^2} \]
where \( Z \) is the atomic number (for hydrogen, \( Z=1 \)), \( R_H \) is the Rydberg constant, and \( n \) is the principal quantum number.

2. **Calculate the Energy Difference**:
The energy difference \( \Delta E \) when the electron transitions from \( n=5 \) to \( n=1 \) is:
\[ \Delta E = E_1 - E_5 = -R_H \left( \frac{1^2}{1^2} - \frac{1^2}{5^2} \right) = -R_H \left( 1 - \frac{1}{25} \right) = -R_H \left( \frac{24}{25} \right) \]

3. **Photon Energy**:
The energy of the emitted photon during this transition is equal to the energy difference:
\[ E_{\text{photon}} = \Delta E = \frac{24 R_H}{25} \]

4. **Relate Energy to Momentum**:
The momentum \( p \) of the emitted photon can be expressed as:
\[ p = \frac{E_{\text{photon}}}{c} = \frac{24 R_H}{25c} \]
where \( c \) is the speed of light.

5. **Conservation of Momentum**:
According to the conservation of momentum, the momentum of the emitted photon is equal to the momentum gained by the hydrogen atom (which has mass \( m \)):
\[ p_{\text{photon}} = m v \]
Therefore:
\[ m v = \frac{24 R_H}{25c} \]

6. **Express Velocity**:
The velocity \( v \) of the hydrogen atom can be expressed as:
\[ v = \frac{24 R_H}{25mc} \]

7. **Relate to Given Expression**:
We are given that the velocity can also be expressed as:
\[ v = k h R \]
To find \( k \), we need to relate \( v \) from both expressions:
\[ k h R = \frac{24 R_H}{25mc} \]

8. **Solve for \( k \)**:
Rearranging gives:
\[ k = \frac{24 R_H}{25m c h} \]

### Final Value of \( k \):
From the above expression, we can identify that:
\[ k = \frac{24}{25} \]

Thus, the value of \( k \) is \( \frac{24}{25} \).