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 solve the question regarding the wavelength (λ) of emitted radiation in the Lyman series, we can follow these steps:

### Step 1: Understanding the Lyman Series
The Lyman series refers to the set of spectral lines corresponding to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 2) to the first energy level (n = 1).

### Step 2: Rydberg Formula
The wavelength of emitted radiation can be derived from the Rydberg formula for hydrogen-like atoms:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)),
- \(n_1\) is the lower energy level (for Lyman series, \(n_1 = 1\)),
- \(n_2\) is the higher energy level (where \(n_2\) can be 2, 3, ...).

### Step 3: Applying the Formula
For the Lyman series, we set \(n_1 = 1\) and \(n_2 = n\) (where \(n\) can be any integer greater than 1). Thus, the formula simplifies to:
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) = R \left( 1 - \frac{1}{n^2} \right) \]

### Step 4: Final Expression for Wavelength
Rearranging the equation gives us:
\[ \lambda = \frac{1}{R \left( 1 - \frac{1}{n^2} \right)} \]
This equation provides the wavelength of the emitted radiation for the Lyman series transitions.