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 quantum 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 emits a photon of wavelength \( \lambda \), we can use the Rydberg formula. Here’s the step-by-step solution:

### Step 1: Write the Rydberg Formula
The Rydberg formula for the wavelengths of spectral lines in hydrogen is given by:

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

where:
- \( \lambda \) is the wavelength of the emitted photon,
- \( R \) is the Rydberg constant,
- \( n_f \) is the final principal quantum number (ground state, \( n_f = 1 \)),
- \( n_i \) is the initial principal quantum number (excited state).

### Step 2: Substitute the Final State
Since the hydrogen atom transitions from an excited state to the ground state, we set \( n_f = 1 \):

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

### Step 3: Simplify the Equation
This simplifies to:

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

### Step 4: Rearrange the Equation
Rearranging the equation gives:

\[ \frac{1}{\lambda} = R - \frac{R}{n_i^2} \]

### Step 5: Isolate \( \frac{1}{n_i^2} \)
Now, isolate \( \frac{1}{n_i^2} \):

\[ \frac{R}{n_i^2} = R - \frac{1}{\lambda} \]

### Step 6: Solve for \( n_i^2 \)
Multiplying both sides by \( n_i^2 \) and rearranging gives:

\[ n_i^2 = \frac{R \lambda}{R \lambda - 1} \]

### Step 7: Take the Square Root
Finally, taking the square root to find \( n_i \):

\[ n_i = \sqrt{\frac{R \lambda}{R \lambda - 1}} \]

### Conclusion
Thus, the principal quantum number \( n \) of the excited state is:

\[ n = \sqrt{\frac{R \lambda}{R \lambda - 1}} \]