An excited state of H atom emits a photon of wavelength `lamda` and returns in the ground state. The principal quantum number of excited state is given by:

Options

A. `sqrt(lamdaR(lamdaR-1))`
B. `sqrt((lamdaR)/((lamdaR-1))`
C. `sqrt(lamdaR(lamdaR-1)`
D. `sqrt((lamdaR-1)/((lamdaR))`

Answer

To find the principal quantum number of the excited state of a hydrogen atom that emits a photon of wavelength \( \lambda \) and returns to the ground state, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Energy Levels**:
- The ground state of hydrogen corresponds to the principal quantum number \( n_1 = 1 \).
- The excited state corresponds to \( n_2 \), which we need to find.

2. **Use the Rydberg Formula**:
- The Rydberg formula for the wavelengths of emitted photons is given by:
\[ \frac{1}{\lambda} = R \cdot \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R \) is the Rydberg constant, and for hydrogen, \( Z = 1 \).

3. **Substitute Known Values**:
- Since \( n_1 = 1 \), we can substitute this into the formula:
\[ \frac{1}{\lambda} = R \cdot \left( 1 - \frac{1}{n_2^2} \right) \]

4. **Rearrange the Equation**:
- Rearranging the equation gives:
\[ \frac{1}{\lambda} = R \cdot \left( 1 - \frac{1}{n_2^2} \right) \]
- This can be rewritten as:
\[ \frac{1}{\lambda} = R - \frac{R}{n_2^2} \]

5. **Isolate \( \frac{1}{n_2^2} \)**:
- Rearranging further, we get:
\[ \frac{R}{n_2^2} = R - \frac{1}{\lambda} \]
- Thus,
\[ \frac{1}{n_2^2} = \frac{R - \frac{1}{\lambda}}{R} \]

6. **Simplify the Expression**:
- This leads to:
\[ n_2^2 = \frac{R \cdot \lambda}{R \cdot \lambda - 1} \]

7. **Take the Square Root**:
- Finally, taking the square root gives us the principal quantum number of the excited state:
\[ n_2 = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]

### Final Result
The principal quantum number of the excited state \( n_2 \) is:
\[ n_2 = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]