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 will follow these steps:

### Step 1: Understand the Series and Transitions
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level. The alpha line of the Lyman series corresponds to the transition from n=2 to n=1. The Paschen series corresponds to transitions where the electron falls to the n=3 energy level. The beta line of the Paschen series corresponds to the transition from n=5 to n=3.

### Step 2: Use the Rydberg Formula
The wavelength of the emitted light during these transitions can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate the Wavelength for the Alpha Line of the Lyman Series
For the alpha line of the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_{\text{Lyman, alpha}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus, we have:
\[ \frac{1}{\lambda_{\text{Lyman, alpha}}} = \frac{3R}{4} \]

### Step 4: Calculate the Wavelength for the Beta Line of the Paschen Series
For the beta line of the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = 5 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_{\text{Paschen, beta}}} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{9} - \frac{1}{25} \right) \]

### Step 5: Simplify the Expression for the Beta Line
To simplify \( \frac{1}{9} - \frac{1}{25} \):
\[ \frac{1}{9} - \frac{1}{25} = \frac{25 - 9}{225} = \frac{16}{225} \]
Thus:
\[ \frac{1}{\lambda_{\text{Paschen, beta}}} = R \left( \frac{16}{225} \right) \]

### Step 6: Relate the Two Wavelengths
Now we have:
1. \( \frac{1}{\lambda_{\text{Lyman, alpha}}} = \frac{3R}{4} \)
2. \( \frac{1}{\lambda_{\text{Paschen, beta}}} = R \left( \frac{16}{225} \right) \)

To find the relationship between the two wavelengths, we can set up the ratio:
\[ \frac{\lambda_{\text{Paschen, beta}}}{\lambda_{\text{Lyman, alpha}}} = \frac{\frac{4}{3R}}{\frac{225}{16R}} = \frac{4 \cdot 16}{3 \cdot 225} = \frac{64}{675} \]

### Step 7: Solve for the Wavelength of the Beta Line
Given that \( \lambda_{\text{Lyman, alpha}} = \lambda \):
\[ \lambda_{\text{Paschen, beta}} = \frac{64}{675} \lambda \]

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