If `lambda_1 and lambda_2` , are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of `lambda_1:lambda_2` is:

Options

A. `1:9`
B. `7:108`
C. `7:135`
D. `1:3`

Answer

To solve the problem, we need to find the ratio of the wavelengths \( \lambda_1 \) and \( \lambda_2 \) for the third member of the Lyman series and the first member of the Paschen series, respectively.

### Step-by-Step Solution:

1. **Identify the Transitions for Lyman Series:**
- The Lyman series corresponds to transitions where the electron falls to the first energy level \( n_1 = 1 \).
- The third member of the Lyman series corresponds to the transition from \( n_2 = 4 \) to \( n_1 = 1 \).

2. **Calculate \( \lambda_1 \) for Lyman Series:**
- The formula for the wavelength in the hydrogen atom is given by:
\[ \frac{1}{\lambda} = R_z \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Here, \( R_z \) is the Rydberg constant for hydrogen, and for our case:
- \( n_1 = 1 \)
- \( n_2 = 4 \)
- Plugging in the values:
\[ \frac{1}{\lambda_1} = R_z \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R_z \left( 1 - \frac{1}{16} \right) = R_z \left( \frac{15}{16} \right) \]
- Therefore, we have:
\[ \lambda_1 = \frac{16}{15 R_z} \]

3. **Identify the Transitions for Paschen Series:**
- The Paschen series corresponds to transitions where the electron falls to the third energy level \( n_1 = 3 \).
- The first member of the Paschen series corresponds to the transition from \( n_2 = 4 \) to \( n_1 = 3 \).

4. **Calculate \( \lambda_2 \) for Paschen Series:**
- Using the same formula:
\[ \frac{1}{\lambda_2} = R_z \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R_z \left( \frac{1}{9} - \frac{1}{16} \right) \]
- Finding a common denominator (which is 144):
\[ \frac{1}{\lambda_2} = R_z \left( \frac{16}{144} - \frac{9}{144} \right) = R_z \left( \frac{7}{144} \right) \]
- Therefore, we have:
\[ \lambda_2 = \frac{144}{7 R_z} \]

5. **Find the Ratio \( \frac{\lambda_1}{\lambda_2} \):**
- Now, we can find the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{16}{15 R_z}}{\frac{144}{7 R_z}} = \frac{16 \cdot 7}{15 \cdot 144} \]
- Simplifying this:
\[ \frac{\lambda_1}{\lambda_2} = \frac{112}{2160} = \frac{7}{135} \]

### Final Answer:
Thus, the value of \( \frac{\lambda_1}{\lambda_2} \) is \( \frac{7}{135} \).