If `lambda_(1) and lambda_(2)` are the wavelength of the first members of the Lyman and Paschen series, respectively , then `lambda_(1) lambda_(2)` is

Options

A. `1:3`
B. `1:30`
C. `7:50`
D. `7:108`

Explanation

To solve the problem, we need to find the product of the wavelengths \( \lambda_1 \) (the first member of the Lyman series) and \( \lambda_2 \) (the first member of the Paschen series) and express it in a specific form.

### Step-by-Step Solution:

1. **Identify the Series and Their Transitions:**
- The Lyman series corresponds to transitions where the final energy level \( n_f = 1 \).
- The first member of the Lyman series corresponds to the transition from \( n_i = 2 \) to \( n_f = 1 \).
- The Paschen series corresponds to transitions where the final energy level \( n_f = 3 \).
- The first member of the Paschen series corresponds to the transition from \( n_i = 4 \) to \( n_f = 3 \).

2. **Use the Rydberg Formula:**
The Rydberg formula for the wavelength of emitted light is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant.

3. **Calculate \( \lambda_1 \) for the Lyman Series:**
- For the first member of the Lyman series:
- \( n_f = 1 \)
- \( n_i = 2 \)
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_1 = \frac{4}{3R} \]

4. **Calculate \( \lambda_2 \) for the Paschen Series:**
- For the first member of the Paschen series:
- \( n_f = 3 \)
- \( n_i = 4 \)
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
To combine the fractions:
\[ \frac{1}{\lambda_2} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda_2 = \frac{144}{7R} \]

5. **Calculate the Product \( \lambda_1 \lambda_2 \):**
\[ \lambda_1 \lambda_2 = \left( \frac{4}{3R} \right) \left( \frac{144}{7R} \right) = \frac{576}{21R^2} \]

6. **Simplify the Expression:**
\[ \lambda_1 \lambda_2 = \frac{192}{7R^2} \]

### Final Result:
The product \( \lambda_1 \lambda_2 \) is given by:
\[ \lambda_1 \lambda_2 = \frac{192}{7R^2} \]