A hydrogen atom in its ground state absorbs `10.2 eV` of energy. The orbital angular momentum is increased by

Options

A. `1.05 xx 10^(-34)J-sec`
B. `3.36 xx 10^(-34)J-sec`
C. `2.11 xx 10^(-34)J-sec`
D. `4.22 xx 10^(-34)J-sec`

Explanation

To solve the problem of how much the orbital angular momentum of a hydrogen atom increases when it absorbs 10.2 eV of energy, we can follow these steps:

### Step-by-Step Solution:

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

2. **Energy Absorption**:
The atom absorbs 10.2 eV of energy. This energy is sufficient to promote the electron from the ground state (n=1) to the first excited state (n=2). The energy levels of the hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \text{ eV}}{n^2} \]
- For \( n = 1 \): \( E_1 = -13.6 \text{ eV} \)
- For \( n = 2 \): \( E_2 = -3.4 \text{ eV} \)

The energy difference between these two states is:
\[ E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV} \]

3. **Determine the Final State**:
After absorbing 10.2 eV, the atom transitions from \( n = 1 \) to \( n = 2 \).

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

- For \( n = 1 \):
\[ L_1 = 1 \cdot \frac{h}{2\pi} = \frac{h}{2\pi} \]

- For \( n = 2 \):
\[ L_2 = 2 \cdot \frac{h}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi} \]

5. **Calculate the Increase in Angular Momentum**:
The increase in angular momentum (\( \Delta L \)) when transitioning from \( n = 1 \) to \( n = 2 \) is:
\[ \Delta L = L_2 - L_1 = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{\pi} - \frac{h}{2\pi} = \frac{h}{2\pi} \]

6. **Final Calculation**:
Substituting \( h \):
\[ \Delta L = \frac{6.626 \times 10^{-34} \text{ J s}}{2\pi} \approx 1.05 \times 10^{-34} \text{ J s} \]

### Conclusion:
The orbital angular momentum of the hydrogen atom increases by \( \frac{h}{2\pi} \) or approximately \( 1.05 \times 10^{-34} \text{ J s} \).

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