A photon of energy `12.09 eV` is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of excited state is

Options

A. 4
B. 5
C. 3
D. 2

Explanation

To solve the problem of determining the quantum number of the excited state of a hydrogen atom after absorbing a photon of energy 12.09 eV, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ground State Energy**:
The energy of the hydrogen atom in the ground state (n=1) is given by:
\[ E_1 = -13.6 \, \text{eV} \]

2. **Energy of the Excited State**:
The energy of the hydrogen atom in the excited state (n) can be expressed as:
\[ E_n = -\frac{13.6}{n^2} \, \text{eV} \]

3. **Energy Difference**:
When the photon is absorbed, the energy difference between the excited state and the ground state can be calculated as:
\[ E_n - E_1 = 12.09 \, \text{eV} \]
Substituting the expressions for \(E_n\) and \(E_1\):
\[ -\frac{13.6}{n^2} - (-13.6) = 12.09 \]
Simplifying this gives:
\[ 13.6 - \frac{13.6}{n^2} = 12.09 \]

4. **Rearranging the Equation**:
Rearranging the equation to isolate the term involving \(n\):
\[ 13.6 - 12.09 = \frac{13.6}{n^2} \]
\[ 1.51 = \frac{13.6}{n^2} \]

5. **Solving for \(n^2\)**:
Now, we can solve for \(n^2\):
\[ n^2 = \frac{13.6}{1.51} \]
Calculating this gives:
\[ n^2 \approx 9.01 \]

6. **Finding the Quantum Number \(n\)**:
Taking the square root to find \(n\):
\[ n \approx 3 \]

### Conclusion:
The quantum number of the excited state is \(n = 3\).