An excited hydrogen atom emits a photon of wavelength `lambda` in returning to the ground state. If 'R' is the Rydberg's constant, then the quantum number 'n' of the excited state is:

Options

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

Answer

To find the quantum number \( n \) of the excited state of a hydrogen atom that emits a photon of wavelength \( \lambda \) while returning to the ground state, we can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

### Step-by-Step Solution:

1. **Identify the Ground State Quantum Number**:
The ground state of the hydrogen atom corresponds to the quantum number \( n_1 = 1 \).

2. **Substitute \( n_1 \) into the Rydberg Formula**:
Substitute \( n_1 = 1 \) into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_2^2} \right) \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \frac{1}{\lambda} = R - \frac{R}{n_2^2} \]
Now, isolate the term involving \( n_2 \):
\[ \frac{R}{n_2^2} = R - \frac{1}{\lambda} \]

4. **Solve for \( \frac{1}{n_2^2} \)**:
Rearranging further gives:
\[ \frac{1}{n_2^2} = \frac{R}{R - \frac{1}{\lambda}} \]

5. **Take the Reciprocal**:
Taking the reciprocal gives:
\[ n_2^2 = \frac{R - \frac{1}{\lambda}}{R} \]

6. **Final Expression for \( n_2 \)**:
Therefore, we can express \( n_2 \) as:
\[ n_2 = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]

### Conclusion:
The quantum number \( n \) of the excited state is given by:
\[ n_2 = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]