During a transition in a hydrogen atom a photon of energy 13.06 eV is emitted. The change in the angular momentum of electron which takes part in this transition is equal to

Options

A. `h/pi`
B. `(2h)/pi`
C. `3/2 h/pi`
D. `5/2 h/mu`

Explanation

To solve the problem of finding the change in angular momentum of an electron in a hydrogen atom during a transition that emits a photon of energy 13.06 eV, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Energy Transition**:
The energy of the emitted photon is given as \( E = 13.06 \, \text{eV} \). In a hydrogen atom, the energy levels are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Determine the Initial and Final Energy Levels**:
The transition occurs from a higher energy level \( n \) to the ground state \( n = 1 \). Therefore, we can write:
\[ \Delta E = E_1 - E_n \]
Given that \( \Delta E = 13.06 \, \text{eV} \), we can express this as:
\[ E_1 - E_n = 13.06 \, \text{eV} \]

3. **Calculate the Energy of the Ground State**:
The energy of the ground state (n=1) is:
\[ E_1 = -13.6 \, \text{eV} \]
Substituting this into the energy difference equation:
\[ -13.6 - E_n = 13.06 \]
Rearranging gives:
\[ E_n = -13.6 - 13.06 = -26.66 \, \text{eV} \]

4. **Find the Principal Quantum Number \( n \)**:
We set \( E_n = -\frac{13.6}{n^2} \):
\[ -\frac{13.6}{n^2} = -26.66 \]
Solving for \( n^2 \):
\[ n^2 = \frac{13.6}{26.66} \approx 0.51 \]
This gives \( n \approx 5 \) (since \( n \) must be a whole number).

5. **Calculate the Angular Momentum**:
The angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
For \( n = 5 \):
\[ L_5 = 5 \frac{h}{2\pi} \]
For \( n = 1 \):
\[ L_1 = 1 \frac{h}{2\pi} \]

6. **Determine the Change in Angular Momentum**:
The change in angular momentum \( \Delta L \) is:
\[ \Delta L = L_5 - L_1 = \left(5 \frac{h}{2\pi}\right) - \left(1 \frac{h}{2\pi}\right) = 4 \frac{h}{2\pi} = 2 \frac{h}{\pi} \]

### Final Answer
The change in angular momentum of the electron during the transition is:
\[ \Delta L = \frac{2h}{\pi} \]