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, we need to find the quantum number of the excited state of a hydrogen atom after it absorbs a photon of energy 12.09 eV. We will use the energy levels of the hydrogen atom and the relationship between energy and quantum numbers.

### Step-by-Step Solution:

1. **Understand the Energy Levels of Hydrogen Atom**:
The energy of an electron in the nth energy level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

2. **Energy of the Ground State**:
For the ground state (n=1), the energy is:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

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

4. **Energy Absorption**:
When the hydrogen atom absorbs a photon of energy 12.09 eV, the energy difference between the ground state and the excited state can be expressed as:
\[ E_1 - E_n = 12.09 \, \text{eV} \]
Substituting the values:
\[ -13.6 \, \text{eV} - \left(-\frac{13.6 \, \text{eV}}{n^2}\right) = 12.09 \, \text{eV} \]

5. **Rearranging the Equation**:
This simplifies to:
\[ -13.6 + \frac{13.6}{n^2} = 12.09 \]
Rearranging gives:
\[ \frac{13.6}{n^2} = 12.09 + 13.6 \]
\[ \frac{13.6}{n^2} = 25.69 \]

6. **Solving for n²**:
Now, we can solve for \( n^2 \):
\[ n^2 = \frac{13.6}{25.69} \]

7. **Calculating n**:
Performing the division:
\[ n^2 \approx 0.529 \]
Taking the square root:
\[ n \approx \sqrt{0.529} \approx 2.3 \]
Since \( n \) must be a whole number, we round up to the nearest whole number:
\[ n \approx 3 \]

8. **Conclusion**:
Therefore, the quantum number of the excited state is \( n = 3 \).

### Final Answer:
The quantum number of the excited state is **3**.