The ratio of the maximum wavlength of the Lyman series in hydrogen specturm to the maximum wavelength in the Panchen series is

Options

A. `(3)/(105)`
B. `(6)(15)`
C. `(52)/(7)`
D. `(7)/(108)`

Answer

To solve the problem of finding the ratio of the maximum wavelength of the Lyman series to the maximum wavelength of the Paschen series in the hydrogen spectrum, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the transitions for the series**:
- The Lyman series corresponds to the transition from \( n = 2 \) to \( n = 1 \).
- The Paschen series corresponds to the transition from \( n = 4 \) to \( n = 3 \).

2. **Use the Rydberg formula**:
The Rydberg formula for the wavelength \( \lambda \) of the emitted light 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.

3. **Calculate for the Lyman series**:
For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = 2 \)
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Therefore, we can express \( \lambda_1 \):
\[ \lambda_1 = \frac{4}{3R} \]

4. **Calculate for the Paschen series**:
For the Paschen series:
- \( n_1 = 3 \)
- \( n_2 = 4 \)
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) \]
To combine these fractions, find a common denominator (which is 144):
\[ \frac{1}{\lambda_2} = R \left( \frac{16}{144} - \frac{9}{144} \right) = R \left( \frac{7}{144} \right) \]
Therefore, we can express \( \lambda_2 \):
\[ \lambda_2 = \frac{144}{7R} \]

5. **Find the ratio of the wavelengths**:
Now, we need to find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{4}{3R}}{\frac{144}{7R}} = \frac{4}{3R} \times \frac{7R}{144} = \frac{4 \times 7}{3 \times 144} = \frac{28}{432} \]
Simplifying this fraction:
\[ \frac{28}{432} = \frac{7}{108} \]

### Final Answer:
The ratio of the maximum wavelength of the Lyman series to the maximum wavelength of the Paschen series is:
\[ \frac{7}{108} \]