Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength `lambda`. If R is the Rydberg constant, then the principal quatum number n of the excited state is

Options

A. `sqrt((lambdaR)/(lambdaR-1))`
B. `sqrt((lambda)/(lambdaR-1))`
C. `sqrt((lambdaR^2)/(lambdaR-1))`
D. `sqrt((lambdaR)/(lambda-1))`

Explanation

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

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

where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the principal quantum number of the lower energy level (ground state),
- \( n_2 \) is the principal quantum number of the higher energy level (excited state).

### Step-by-Step Solution:

1. **Identify the quantum numbers**: Since the atom is transitioning from an excited state to the ground state, we set:
- \( n_1 = 1 \) (ground state)
- \( n_2 = n \) (excited state)

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

3. **Rearrange the equation**: We can rearrange the equation to isolate \( \frac{1}{n^2} \):
\[ \frac{1}{\lambda} = R - \frac{R}{n^2} \]
Rearranging gives:
\[ \frac{R}{n^2} = R - \frac{1}{\lambda} \]

4. **Express \( \frac{1}{n^2} \)**: Now, we can write:
\[ \frac{1}{n^2} = \frac{R}{R - \frac{1}{\lambda}} \]

5. **Take the reciprocal**: To find \( n^2 \), we take the reciprocal of both sides:
\[ n^2 = \frac{R - \frac{1}{\lambda}}{R} \]

6. **Simplify the expression**: This can be rewritten as:
\[ n^2 = \frac{R \lambda - 1}{R \lambda} \]

7. **Find \( n \)**: Finally, taking the square root gives:
\[ n = \sqrt{\frac{R \lambda}{R \lambda - 1}} \]

### Final Result:
Thus, the principal quantum number \( n \) of the excited state is given by:
\[ n = \sqrt{\frac{R \lambda}{R \lambda - 1}} \]