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

Explanation

To find the principal quantum number \( n \) of a hydrogen atom in an excited state that emits a photon of wavelength \( \lambda \) when it returns to the ground state, we can use the Rydberg formula for hydrogen:

### Step-by-Step Solution:

1. **Understanding the Energy of the Photon**:
The energy of the emitted photon can be expressed using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Using the Rydberg Formula**:
The energy difference between two energy levels in a hydrogen atom is given by:
\[ E = R_H \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \]
where \( R_H \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)) and \( n \) is the principal quantum number of the excited state.

3. **Equating the Two Energy Expressions**:
Set the two expressions for energy equal to each other:
\[ \frac{hc}{\lambda} = R_H \left( 1 - \frac{1}{n^2} \right) \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ 1 - \frac{1}{n^2} = \frac{hc}{\lambda R_H} \]

5. **Solving for \( \frac{1}{n^2} \)**:
From the above equation, we can isolate \( \frac{1}{n^2} \):
\[ \frac{1}{n^2} = 1 - \frac{hc}{\lambda R_H} \]

6. **Finding \( n^2 \)**:
Taking the reciprocal:
\[ n^2 = \frac{1}{1 - \frac{hc}{\lambda R_H}} \]

7. **Calculating \( n \)**:
Finally, take the square root to find \( n \):
\[ n = \sqrt{\frac{1}{1 - \frac{hc}{\lambda R_H}}} \]

### Final Answer:
The value of \( n \) is given by:
\[ n = \sqrt{\frac{1}{1 - \frac{hc}{\lambda R_H}}} \]