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 when it absorbs 10.2 eV of energy, we can follow these steps:

### Step 1: Understand the Initial State
The hydrogen atom is initially in the ground state, which corresponds to the principal quantum number \( n_1 = 1 \).

### Step 2: Determine the Energy Absorbed
The energy absorbed by the hydrogen atom is given as \( E = 10.2 \, \text{eV} \).

### Step 3: Calculate the Final State
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
To find the final state \( n_2 \) after absorbing 10.2 eV, we set up the equation:
\[ E_{n_2} - E_{n_1} = 10.2 \, \text{eV} \]
Substituting the energy levels:
\[ -\frac{13.6}{n_2^2} + 13.6 = 10.2 \]
This simplifies to:
\[ -\frac{13.6}{n_2^2} = 10.2 - 13.6 \]
\[ -\frac{13.6}{n_2^2} = -3.4 \]
\[ \frac{13.6}{n_2^2} = 3.4 \]
Now, solving for \( n_2^2 \):
\[ n_2^2 = \frac{13.6}{3.4} = 4 \]
Thus, \( n_2 = 2 \).

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

The change in orbital angular momentum \( \Delta L \) when the electron transitions from \( n_1 \) to \( n_2 \) is:
\[ \Delta L = L_{n_2} - L_{n_1} = \left( n_2 \frac{h}{2\pi} \right) - \left( n_1 \frac{h}{2\pi} \right) \]
Substituting the values of \( n_1 \) and \( n_2 \):
\[ \Delta L = \left( 2 \frac{h}{2\pi} \right) - \left( 1 \frac{h}{2\pi} \right) \]
\[ \Delta L = \frac{h}{2\pi} \]

### Step 5: Substitute Planck's Constant
Using \( h = 6.63 \times 10^{-34} \, \text{J s} \):
\[ \Delta L = \frac{6.63 \times 10^{-34}}{2\pi} \]
Calculating this gives:
\[ \Delta L \approx \frac{6.63 \times 10^{-34}}{6.28} \approx 1.05 \times 10^{-34} \, \text{J s} \]

### Final Answer
The increase in the orbital angular momentum of the electron is approximately \( 1.05 \times 10^{-34} \, \text{J s} \).

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