When a hydrogen atom emits a photon in going from n=5 to n=1, its recoil speed is almost

Options

A. `10^(-4) m//sec`
B. `2xx10^(-2) m//sec`
C. `4 m//sec`
D. `8xx10^(2) m//sec`

Explanation

To find the recoil speed of a hydrogen atom when it emits a photon while transitioning from n=5 to n=1, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Conservation of Momentum**:
When the hydrogen atom emits a photon, the momentum of the hydrogen atom must equal the momentum of the emitted photon. This can be expressed as:
\[ m v = p_{\text{photon}} \]
where \( m \) is the mass of the hydrogen atom, \( v \) is the recoil speed of the hydrogen atom, and \( p_{\text{photon}} \) is the momentum of the photon.

2. **Photon Momentum**:
The momentum of a photon can be expressed in terms of its wavelength \( \lambda \):
\[ p_{\text{photon}} = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

3. **Calculate the Wavelength**:
To find \( \lambda \), we can use the Rydberg formula for the transition from \( n=5 \) to \( n=1 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Here, \( R \) is the Rydberg constant, \( n_1 = 1 \), and \( n_2 = 5 \):
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{25} \right) = R \left( \frac{24}{25} \right) \]

4. **Substituting Rydberg Constant**:
The value of the Rydberg constant \( R \) is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{24}{25} \]

5. **Calculate \( \lambda \)**:
From the above equation, we can find \( \lambda \):
\[ \lambda = \frac{25}{24 \times 1.097 \times 10^7} \]

6. **Substituting into Momentum Equation**:
Now we can substitute \( \lambda \) back into the momentum equation:
\[ m v = \frac{h}{\lambda} \]
Rearranging gives:
\[ v = \frac{h}{m \lambda} \]

7. **Substituting Known Values**:
We know:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Substitute \( \lambda \) from the previous step.

8. **Final Calculation**:
After substituting all values, we can calculate \( v \):
\[ v = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times \lambda} \]
This will yield a recoil speed of approximately \( 4 \, \text{m/s} \).

### Final Answer:
The recoil speed of the hydrogen atom when it emits a photon while transitioning from n=5 to n=1 is approximately **4 m/s**.