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

Options

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

Answer

To find the ratio of the wavelengths of the first members of the Lyman and Paschen series, we will use Rydberg's formula for hydrogen spectral lines.

### Step-by-Step Solution:

**Step 1: Identify the first members of the Lyman and Paschen series.**
- The first member of the Lyman series corresponds to the transition from n=2 to n=1.
- The first member of the Paschen series corresponds to the transition from n=4 to n=3.

**Step 2: Write Rydberg's formula.**
- Rydberg's formula 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 \( \lambda_1 \) for the Lyman series.**
- For the Lyman series, \( n_1 = 1 \) and \( n_2 = 2 \):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore,
\[ \lambda_1 = \frac{4}{3R} \]

**Step 4: Calculate \( \lambda_2 \) for the Paschen series.**
- For the Paschen series, \( n_1 = 3 \) and \( n_2 = 4 \):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
- To combine these fractions, find a common denominator (144):
\[ \frac{1}{\lambda_2} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]
- Therefore,
\[ \lambda_2 = \frac{144}{7R} \]

**Step 5: Find the ratio \( \frac{\lambda_1}{\lambda_2} \).**
- Now, we can find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{4}{3R}}{\frac{144}{7R}} = \frac{4}{3R} \cdot \frac{7R}{144} = \frac{4 \cdot 7}{3 \cdot 144} \]
- Simplifying this gives:
\[ \frac{28}{432} = \frac{7}{108} \]

**Final Answer:**
- The ratio \( \lambda_1 : \lambda_2 = 7 : 108 \).