When an electron in the hydrogen atom in ground state absorb a photon of energy `12.1eV`, its angular momentum

Options

A. decreases by `2.11xx10^(-34) J-s`
B. decreases by `1.055xx10^(-34) J-s`
C. increase by `2.11xx10^(-34) J-s`
D. increase by `1.055xx10^(-34) J-s`

Answer

To solve the problem, we need to determine the change in angular momentum of an electron in a hydrogen atom when it absorbs a photon of energy 12.1 eV.

### Step-by-Step Solution:

1. **Identify the Initial State**:
The electron is in the ground state of the hydrogen atom, which corresponds to the principal quantum number \( n_1 = 1 \).

2. **Photon Energy Absorption**:
The electron absorbs a photon with energy \( E = 12.1 \, \text{eV} \). This energy will cause the electron to transition to a higher energy level.

3. **Energy Levels of Hydrogen Atom**:
The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

4. **Calculate the Final State**:
We need to find the final state \( n_2 \) after the absorption of the photon. The energy difference between the two states can be expressed as:
\[ E = E_{n_1} - E_{n_2} = 12.1 \, \text{eV} \]
Substituting the energy levels:
\[ -\frac{13.6}{1^2} - \left(-\frac{13.6}{n_2^2}\right) = 12.1 \]
This simplifies to:
\[ 13.6 - \frac{13.6}{n_2^2} = 12.1 \]
Rearranging gives:
\[ \frac{13.6}{n_2^2} = 1.5 \]
Thus:
\[ n_2^2 = \frac{13.6}{1.5} \approx 9.0667 \implies n_2 \approx 3 \]

5. **Calculate Angular Momentum**:
The angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L_n = \frac{n h}{2\pi} \]
where \( h \) is Planck's constant (\( h \approx 6.626 \times 10^{-34} \, \text{J s} \)).

- For \( n_1 = 1 \):
\[ L_1 = \frac{1 \cdot h}{2\pi} = \frac{h}{2\pi} \]
- For \( n_2 = 3 \):
\[ L_3 = \frac{3 \cdot h}{2\pi} = \frac{3h}{2\pi} \]

6. **Change in Angular Momentum**:
The change in angular momentum \( \Delta L \) is:
\[ \Delta L = L_3 - L_1 = \frac{3h}{2\pi} - \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \]

7. **Substituting Planck's Constant**:
Now substituting the value of \( h \):
\[ \Delta L = \frac{6.626 \times 10^{-34}}{\pi} \approx \frac{6.626 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \, \text{J s} \]

8. **Conclusion**:
The angular momentum of the electron increases by \( 2.11 \times 10^{-34} \, \text{J s} \).

### Final Answer:
The angular momentum increases by \( 2.11 \times 10^{-34} \, \text{J s} \).