If the wavelength of `alpha`-line of Lyman series in hydrogen atom is `lambda`, find the wavelength of `beta`- line of Paschen series.

Options

A. `(3^(2)xx5^(6))/(2^(3))xxlambda`
B. `(3^(6)xx5^(3))/(2^(2))xx lambda`
C. `(3^(3)xx5^(2))/(2^(6))xx lambda`
D. `(3^(2)xx5^(3))/(2^(6))xx lambda`

Explanation

To solve the problem of finding the wavelength of the beta line of the Paschen series given the wavelength of the alpha line of the Lyman series in a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the transitions for the Lyman series:**
- The Lyman series corresponds to transitions that end at the ground state (n=1).
- The alpha line of the Lyman series corresponds to a transition from n=2 to n=1.

2. **Write the formula for the wavelength:**
- The wavelength (\( \lambda \)) is related to the wave number (\( \bar{\nu} \)) by the formula:
\[ \bar{\nu} = \frac{1}{\lambda} \]
- The wave number can also be expressed using the Rydberg formula:
\[ \bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- Where \( R \) is the Rydberg constant, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

3. **Apply the formula for the alpha line of the Lyman series:**
- For the alpha line (n=2 to n=1):
\[ \bar{\nu}_{\text{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
- Therefore, we have:
\[ \frac{1}{\lambda} = R \left( \frac{3}{4} \right) \]
- Thus, we can express \( \lambda \) as:
\[ \lambda = \frac{4}{3R} \]

4. **Identify the transitions for the Paschen series:**
- The Paschen series corresponds to transitions that end at n=3.
- The beta line of the Paschen series corresponds to a transition from n=5 to n=3.

5. **Apply the formula for the beta line of the Paschen series:**
- For the beta line (n=5 to n=3):
\[ \bar{\nu}_{\text{Paschen}} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{9} - \frac{1}{25} \right) \]
- Finding a common denominator (which is 225):
\[ \frac{1}{9} = \frac{25}{225}, \quad \frac{1}{25} = \frac{9}{225} \]
- Thus:
\[ \bar{\nu}_{\text{Paschen}} = R \left( \frac{25 - 9}{225} \right) = R \left( \frac{16}{225} \right) \]
- Therefore:
\[ \frac{1}{\lambda_{\text{Paschen}}} = R \left( \frac{16}{225} \right) \]
- Hence:
\[ \lambda_{\text{Paschen}} = \frac{225}{16R} \]

6. **Relate the two wavelengths:**
- We have:
\[ \lambda_{\text{Paschen}} = \frac{225}{16R} \]
- And from the Lyman series:
\[ \lambda = \frac{4}{3R} \]
- To express \( \lambda_{\text{Paschen}} \) in terms of \( \lambda \):
\[ \lambda_{\text{Paschen}} = \frac{225}{16} \cdot \frac{3}{4} \cdot \lambda = \frac{675}{64} \lambda \]

### Final Answer:
The wavelength of the beta line of the Paschen series is:
\[ \lambda_{\text{Paschen}} = \frac{675}{64} \lambda \]