A hydrogen atom in ground state absorbs 12.09 eV of energy . The orbital angular momentum of the electron

Options

A. `(3h)/pi`
B. `(2h)/pi`
C. `h/pi`
D. `(4h)/pi`

Answer

To solve the problem, we need to determine the orbital angular momentum of an electron in a hydrogen atom after it absorbs 12.09 eV of energy. Here’s a step-by-step solution:

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

### Step 2: Calculate the energy absorbed
The energy absorbed by the hydrogen atom is given as \( 12.09 \, \text{eV} \).

### Step 3: Use the energy level formula
The energy levels of a hydrogen atom are given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 4: Set up the energy absorption equation
When the atom absorbs energy, it transitions from the initial state \( n_1 \) to a final state \( n_2 \). The change in energy can be expressed as:
\[ \Delta E = E_{n_2} - E_{n_1} \]
Substituting the energy level formula:
\[ \Delta E = \left(-\frac{13.6}{n_2^2}\right) - \left(-\frac{13.6}{1^2}\right) = -\frac{13.6}{n_2^2} + 13.6 \]
Thus,
\[ \Delta E = 13.6 - \frac{13.6}{n_2^2} \]

### Step 5: Substitute the absorbed energy
Since the atom absorbs \( 12.09 \, \text{eV} \), we set up the equation:
\[ 12.09 = 13.6 - \frac{13.6}{n_2^2} \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ \frac{13.6}{n_2^2} = 13.6 - 12.09 \]
\[ \frac{13.6}{n_2^2} = 1.51 \]

### Step 7: Solve for \( n_2^2 \)
Now, we can solve for \( n_2^2 \):
\[ n_2^2 = \frac{13.6}{1.51} \approx 9 \]

### Step 8: Find \( n_2 \)
Taking the square root gives:
\[ n_2 = 3 \]

### Step 9: Calculate the orbital angular momentum
The orbital angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
Substituting \( n = n_2 = 3 \):
\[ L = 3 \frac{h}{2\pi} \]

### Final Answer
The orbital angular momentum of the electron after absorbing the energy is:
\[ L = \frac{3h}{2\pi} \]

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