In Lyman series the wavelength lambda of emitted radiation is given by (R is rydberg constant)

Options

A. `1/lambda=R(1/2^2-1/n^2), n=3,4,5,.....`
B. `1/lambda=R(1/3^2-1/n^2), n=4,5,6,.....`
C. `1/lambda=R(1/1^2-1/n^2), n=2,3,...`
D. `1/lambda=R(1/4^2-1/n^2), n=5,6,7,...`

Answer

To find the wavelength \( \lambda \) of emitted radiation in the Lyman series, we can use the Rydberg formula for hydrogen-like atoms. The Lyman series corresponds to transitions where the final energy level \( n_f \) is 1 (the ground state), and the initial energy level \( n_i \) can be any integer greater than 1 (2, 3, 4, ...).

### Step-by-step Solution:

1. **Understand the Rydberg Formula**: The general formula for the wavelength of emitted radiation is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant, \( n_f \) is the final energy level, and \( n_i \) is the initial energy level.

2. **Identify the Values for Lyman Series**: In the Lyman series:
- The final energy level \( n_f = 1 \) (ground state).
- The initial energy level \( n_i \) can be 2, 3, 4, etc. (any integer greater than 1).

3. **Substitute the Values into the Formula**: For the Lyman series, we substitute \( n_f = 1 \) into the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_i^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_i^2} \right) \]

4. **Express the Result**: This is the expression for the wavelength \( \lambda \) in the Lyman series:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_i^2} \right) \]

### Conclusion:
The wavelength of emitted radiation in the Lyman series can be expressed as:
\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n_i^2} \right) \]
where \( n_i \) can take values 2, 3, 4, etc.