The Wavelength of first member of Balmer series in hydrogen spectrum is `lambda` . Calculate the wavelength of first member of Lymen series in the same spectrum

Options

A. `(5/27) lambda`
B. `(4/27) lambda`
C. `(27/5) lambda`
D. `(27/4) lambda`

Answer

To solve the problem, we need to calculate the wavelength of the first member of the Lyman series in the hydrogen spectrum, given that the wavelength of the first member of the Balmer series is denoted as \( \lambda \).

### Step-by-Step Solution:

1. **Identify the Wavelength of the First Member of the Balmer Series:**
- The first member of the Balmer series corresponds to the transition from \( n_i = 3 \) to \( n_f = 2 \).
- We denote the wavelength of this transition as \( \lambda_1 = \lambda \).

2. **Use the Rydberg Formula for the Balmer Series:**
- The Rydberg formula is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
- For the first member of the Balmer series:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- Simplifying this:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
- Finding a common denominator (36):
\[ \frac{1}{\lambda_1} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
- Therefore:
\[ \lambda_1 = \frac{36}{5R} \quad \text{(Equation 1)} \]

3. **Calculate the Wavelength of the First Member of the Lyman Series:**
- The first member of the Lyman series corresponds to the transition from \( n_i = 2 \) to \( n_f = 1 \).
- We denote the wavelength of this transition as \( \lambda_2 \).
- Using the Rydberg formula for the Lyman series:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
- Simplifying this:
\[ \frac{1}{\lambda_2} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore:
\[ \lambda_2 = \frac{4}{3R} \quad \text{(Equation 2)} \]

4. **Relate \( \lambda_2 \) to \( \lambda_1 \):**
- Now, we can find the ratio of \( \lambda_2 \) to \( \lambda_1 \):
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \cdot \frac{5}{36} \]
- Simplifying this:
\[ \frac{\lambda_2}{\lambda_1} = \frac{20}{108} = \frac{5}{27} \]
- Thus:
\[ \lambda_2 = \frac{5}{27} \lambda_1 \]

5. **Final Result:**
- Since \( \lambda_1 = \lambda \):
\[ \lambda_2 = \frac{5}{27} \lambda \]

### Conclusion:
The wavelength of the first member of the Lyman series in the hydrogen spectrum is \( \frac{5}{27} \lambda \).