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 of finding the ratio of the wavelengths of the first members of the Lyman and Paschen series, we can follow these steps:

### Step 1: Identify the transitions for the Lyman and Paschen series
- The Lyman series corresponds to transitions from higher energy levels to the first energy level (n=1). The first member of the Lyman series is the transition from n=2 to n=1.
- The Paschen series corresponds to transitions from higher energy levels to the third energy level (n=3). The first member of the Paschen series is the transition from n=4 to n=3.

### Step 2: Use the Rydberg formula for wavelength
The Rydberg formula for the wavelength of light emitted during an electron transition 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 (transition from n=2 to n=1):
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]

Thus,

\[ \lambda_1 = \frac{4}{3R} \]

### Step 4: Calculate \( \lambda_2 \) for the Paschen series
For the Paschen series (transition from n=4 to n=3):
- \( n_1 = 3 \)
- \( n_2 = 4 \)

Using the Rydberg formula:

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]

Finding a common denominator (144):

\[ \frac{1}{\lambda_2} = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right) \]

Thus,

\[ \lambda_2 = \frac{144}{7R} \]

### Step 5: Find the ratio \( \lambda_1 : \lambda_2 \)
Now we can find the ratio \( \lambda_1 : \lambda_2 \):

\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{4}{3R}}{\frac{144}{7R}} = \frac{4 \cdot 7}{3 \cdot 144} = \frac{28}{432} = \frac{7}{108} \]

### Final Result
Thus, the ratio of the wavelengths of the first members of the Lyman and Paschen series is:

\[ \lambda_1 : \lambda_2 = \frac{7}{108} \]