A hydrogen atom ia in excited state of principal quantum number `n` . It emits a photon of wavelength `lambda` when it returnesto the ground state. The value of `n` is

Options

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

Answer

To solve the problem, we need to determine the principal quantum number \( n \) of a hydrogen atom that emits a photon of wavelength \( \lambda \) when it transitions from an excited state to the ground state.

### Step-by-Step Solution:

1. **Understand the Transition**:
The hydrogen atom transitions from an excited state with principal quantum number \( n \) to the ground state, which has \( n_1 = 1 \).

2. **Use the Wavelength Formula**:
The wavelength of the emitted photon can be described by the formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level (ground state, \( n_1 = 1 \)), and \( n_2 \) is the higher energy level (excited state, \( n_2 = n \)).

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

4. **Rearranging the Equation**:
Rearranging gives us:
\[ \frac{1}{\lambda} = R - \frac{R}{n^2} \]
This can be rewritten as:
\[ \frac{R}{n^2} = R - \frac{1}{\lambda} \]

5. **Isolate \( n^2 \)**:
Rearranging further, we find:
\[ \frac{1}{n^2} = \frac{R - \frac{1}{\lambda}}{R} \]
Taking the common denominator on the right-hand side gives:
\[ \frac{1}{n^2} = \frac{\lambda R - 1}{\lambda R} \]

6. **Taking Reciprocals**:
Taking the reciprocal of both sides, we get:
\[ n^2 = \frac{\lambda R}{\lambda R - 1} \]

7. **Finding \( n \)**:
Finally, taking the square root of both sides yields:
\[ n = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]

### Conclusion:
Thus, the value of \( n \) is:
\[ n = \sqrt{\frac{\lambda R}{\lambda R - 1}} \]