The wavelength of first member of Balmer series is `6563 Å`. Calculate the wavelength of second member of Lyman series.

Options

A. 1025.5 A
B. 2050 A
C. 6563 A
D. none of these

Answer

To solve the problem of calculating the wavelength of the second member of the Lyman series, given that the wavelength of the first member of the Balmer series is 6563 Å, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Series**:
- The Balmer series corresponds to transitions from higher energy levels (n = 3, 4, 5, ...) to n = 2.
- The Lyman series corresponds to transitions from higher energy levels (n = 2, 3, 4, ...) to n = 1.

2. **Identify the First Member of the Balmer Series**:
- The first member of the Balmer series corresponds to the transition from n = 3 to n = 2.
- The given wavelength (λ1) of the first member of the Balmer series is 6563 Å.

3. **Use the Rydberg Formula**:
- The Rydberg formula for the wavelength of emitted light is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
- For the first member of the Balmer series (n_i = 3, n_f = 2):
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- This can be simplified to:
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

4. **Calculate the Wavelength of the Second Member of the Lyman Series**:
- The second member of the Lyman series corresponds to the transition from n = 3 to n = 1.
- Using the Rydberg formula for the second member of the Lyman series:
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
- This can be simplified to:
\[ \frac{1}{\lambda_2} = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right) \]

5. **Relate the Two Wavelengths**:
- We can relate λ1 and λ2 using the ratios of their respective terms:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\left( \frac{8}{9} \right)}{\left( \frac{5}{36} \right)} \]
- Rearranging gives:
\[ \lambda_2 = \lambda_1 \cdot \frac{8}{9} \cdot \frac{36}{5} \]

6. **Substituting the Value of λ1**:
- Substitute λ1 = 6563 Å into the equation:
\[ \lambda_2 = 6563 \cdot \frac{8 \cdot 36}{9 \cdot 5} \]
- Calculate:
\[ \lambda_2 = 6563 \cdot \frac{288}{45} = 6563 \cdot 6.4 \]

7. **Final Calculation**:
- Performing the final multiplication:
\[ \lambda_2 \approx 1025.5 \text{ Å} \]

### Final Answer:
The wavelength of the second member of the Lyman series is approximately **1025.5 Å**.