When a hydrogen atoms emits a photon of energy `12.1 eV` , its orbital angular momentum changes by (where h os Planck's constant)

Options

A. `3h/pi`
B. `2h/pi`
C. `h/pi`
D. `4h/pi`

Answer

To solve the problem of how the orbital angular momentum of a hydrogen atom changes when it emits a photon of energy 12.1 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Energy Transition**:
The energy of the emitted photon is given as 12.1 eV. In hydrogen, this energy corresponds to a transition between energy levels. Specifically, this energy corresponds to the transition from the n=3 level to the n=1 level.

2. **Determine the Initial and Final Quantum Numbers**:
For the transition:
- Initial quantum number \( n_1 = 3 \)
- Final quantum number \( n_2 = 1 \)

3. **Calculate the Change in Quantum Number**:
The change in the principal quantum number \( n \) is given by:
\[ \Delta n = n_1 - n_2 = 3 - 1 = 2 \]

4. **Use the Formula for Orbital Angular Momentum**:
The orbital angular momentum \( L \) of an electron in a hydrogen atom is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( h \) is Planck's constant.

5. **Calculate the Initial and Final Orbital Angular Momentum**:
- Initial orbital angular momentum when \( n = 3 \):
\[ L_1 = 3 \frac{h}{2\pi} \]
- Final orbital angular momentum when \( n = 1 \):
\[ L_2 = 1 \frac{h}{2\pi} \]

6. **Determine the Change in Orbital Angular Momentum**:
The change in orbital angular momentum \( \Delta L \) is given by:
\[ \Delta L = L_2 - L_1 = \left(1 \frac{h}{2\pi}\right) - \left(3 \frac{h}{2\pi}\right) \]
Simplifying this gives:
\[ \Delta L = \frac{h}{2\pi} - \frac{3h}{2\pi} = -\frac{2h}{2\pi} = -\frac{h}{\pi} \]

7. **Conclusion**:
The change in the orbital angular momentum when the hydrogen atom emits a photon of energy 12.1 eV is:
\[ \Delta L = -\frac{h}{\pi} \]