The first line in the Lyman series has wavelength `lambda`. The wavelength of the first line in Balmer series is

Options

A. `(2)/(9)lambda`
B. `(9)/(2)lambda`
C. `(5)/(27)lambda`
D. `(27)/(5)lambda`

Explanation

To solve the problem of finding the wavelength of the first line in the Balmer series given that the first line in the Lyman series has a wavelength of `lambda`, we can follow these steps:

### Step 1: Understand the Lyman and Balmer Series
The Lyman series corresponds to transitions where electrons fall to the n=1 energy level from higher levels (n=2, 3, 4,...). The first line in the Lyman series corresponds to the transition from n=2 to n=1.

The Balmer series corresponds to transitions where electrons fall to the n=2 energy level from higher levels (n=3, 4, 5,...). The first line in the Balmer series corresponds to the transition from n=3 to n=2.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light during electron transitions is given by:

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

where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

### Step 3: Calculate Wavelength for the Lyman Series
For the first line in the Lyman series (n=2 to n=1):

\[ \frac{1}{\lambda_{\text{Lyman}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus, we can express the wavelength as:

\[ \lambda_{\text{Lyman}} = \frac{4}{3R} \]

### Step 4: Calculate Wavelength for the Balmer Series
For the first line in the Balmer series (n=3 to n=2):

\[ \frac{1}{\lambda_{\text{Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda_{\text{Balmer}}} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]

Thus, we can express the wavelength as:

\[ \lambda_{\text{Balmer}} = \frac{36}{5R} \]

### Step 5: Relate the Two Wavelengths
From the previous calculations, we have:

\[ \lambda_{\text{Lyman}} = \frac{4}{3R} \quad \text{and} \quad \lambda_{\text{Balmer}} = \frac{36}{5R} \]

Now, we can relate the two wavelengths:

\[ \frac{\lambda_{\text{Balmer}}}{\lambda_{\text{Lyman}}} = \frac{\frac{36}{5R}}{\frac{4}{3R}} = \frac{36 \cdot 3}{5 \cdot 4} = \frac{108}{20} = \frac{27}{5} \]

Thus, we can express the wavelength of the Balmer series in terms of the Lyman series:

\[ \lambda_{\text{Balmer}} = \frac{27}{5} \lambda_{\text{Lyman}} \]

### Final Answer
Since the first line in the Lyman series has a wavelength of `lambda`, we can write:

\[ \lambda_{\text{Balmer}} = \frac{27}{5} \lambda \]