If the wavelength of the first member of Lyman series of hydrogen is `lambda` .The wavelength of the second member will be

Options

A. `(27)/(5)lambda`
B. `(32)/(27)lambda`
C. `(27)/(32)lambda`
D. `(5)/(27)lambda`

Answer

To find the wavelength of the second member of the Lyman series of hydrogen, we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 2) to the first energy level (n = 1). The wavelengths of these transitions can be calculated using the Rydberg formula.

### Step 2: Write the Rydberg Formula
The Rydberg formula for the wavelength (λ) of the emitted light is given by:

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

Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate the Wavelength for the First Member of the Lyman Series
For the first member of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the Rydberg formula:

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

Calculating the right-hand side:

\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus:

\[ \lambda = \frac{4}{3R} \]

### Step 4: Calculate the Wavelength for the Second Member of the Lyman Series
For the second member of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 3 \)

Substituting these values into the Rydberg formula:

\[ \frac{1}{\lambda'} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]

Calculating the right-hand side:

\[ \frac{1}{\lambda'} = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]

Thus:

\[ \lambda' = \frac{9}{8R} \]

### Step 5: Find the Relationship Between λ and λ'
Now, we have:

1. \( \lambda = \frac{4}{3R} \)
2. \( \lambda' = \frac{9}{8R} \)

To find the relationship between \( \lambda' \) and \( \lambda \), we can express \( \lambda' \) in terms of \( \lambda \):

\[ \lambda' = \frac{9}{8R} = \frac{9}{8} \cdot \frac{R}{4} \cdot 3 = \frac{27}{32} \lambda \]

### Final Answer
The wavelength of the second member of the Lyman series is:

\[ \lambda' = \frac{27}{32} \lambda \]