A hydrogen atom in ground state absorbs `10.2eV` of energy .The orbital angular momentum of the electron is increases by

Options

A. `1.05 xx 10^(-34)J s`
B. `2.11 xx 10^(-34)J s`
C. `3.16 xx 10^(-34)J s`
D. `4.22 xx 10^(-34)J s`

Answer

To solve the problem of how much the orbital angular momentum of the electron in a hydrogen atom increases after absorbing 10.2 eV of energy, we can follow these steps:

### Step 1: Identify the initial state of the hydrogen atom
The hydrogen atom is in the ground state, which corresponds to the principal quantum number \( n_1 = 1 \).

### Step 2: Determine the energy levels of the hydrogen atom
The energy of the electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the ground state (\( n_1 = 1 \)):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

### Step 3: Calculate the energy after absorption
When the atom absorbs 10.2 eV of energy, the total energy becomes:
\[ E_{\text{final}} = E_1 + 10.2 \, \text{eV} = -13.6 \, \text{eV} + 10.2 \, \text{eV} = -3.4 \, \text{eV} \]

### Step 4: Find the new principal quantum number \( n_2 \)
We need to find \( n_2 \) such that:
\[ E_{n_2} = -\frac{13.6 \, \text{eV}}{n_2^2} = -3.4 \, \text{eV} \]
Setting the equations equal gives:
\[ -\frac{13.6}{n_2^2} = -3.4 \]
Solving for \( n_2^2 \):
\[ \frac{13.6}{3.4} = n_2^2 \implies n_2^2 = 4 \implies n_2 = 2 \]

### Step 5: Calculate the change in orbital angular momentum
The orbital angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
For the ground state (\( n_1 = 1 \)):
\[ L_1 = 1 \cdot \frac{h}{2\pi} \]
For the excited state (\( n_2 = 2 \)):
\[ L_2 = 2 \cdot \frac{h}{2\pi} \]

### Step 6: Find the change in orbital angular momentum
The change in orbital angular momentum \( \Delta L \) is:
\[ \Delta L = L_2 - L_1 = \left(2 \cdot \frac{h}{2\pi}\right) - \left(1 \cdot \frac{h}{2\pi}\right) = \frac{h}{2\pi} \]

### Step 7: Substitute the value of \( h \)
Using \( h \approx 6.626 \times 10^{-34} \, \text{Js} \):
\[ \Delta L = \frac{6.626 \times 10^{-34}}{2\pi} \approx \frac{6.626 \times 10^{-34}}{6.2832} \approx 1.055 \times 10^{-34} \, \text{Js} \]

### Conclusion
The orbital angular momentum of the electron increases by approximately \( 1.055 \times 10^{-34} \, \text{Js} \).