If `lamda_1 " and " lamda_2` are the wavelengths of the first members of the Lyman and paschen series respectively, then `lamda_1/lamda_2` is equal to

Options

A. (a)`1:3`
B. (b)`1:30`
C. (c)`7:50`
D. (d)`7:108`

Answer

To solve the problem of finding the ratio of the wavelengths \( \frac{\lambda_1}{\lambda_2} \) for the first members of the Lyman and Paschen series, we will follow these steps:

### Step 1: Understand the Series and Identify the Values
- The Lyman series corresponds to transitions where the electron falls to the first energy level (n=1) from higher levels (n=2, 3, 4,...).
- The first member of the Lyman series is the transition from n=2 to n=1.
- The Paschen series corresponds to transitions where the electron falls to the third energy level (n=3) from higher levels (n=4, 5, 6,...).
- The first member of the Paschen series is the transition from n=4 to n=3.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of the emitted photon during these transitions 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.

### Step 3: Calculate \( \lambda_1 \) for the Lyman Series
For the first member of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

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, we have:

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

### Step 4: Calculate \( \lambda_2 \) for the Paschen Series
For the first member of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = 4 \)

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, we have:

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

### Step 5: Calculate the Ratio \( \frac{\lambda_1}{\lambda_2} \)
Now we find the ratio of the two wavelengths:

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

The \( R \) cancels out:

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

Simplifying \( \frac{28}{432} \):

\[ \frac{28}{432} = \frac{7}{108} \]

### Final Answer
Thus, the ratio \( \frac{\lambda_1}{\lambda_2} \) is:

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