The ratio of difference in wavelengths of `1^(st)` and `2^(nd)` lines of Lyman series in H-like atom to difference in wavelength for `2^(nd)` and `3^(rd)` lines of same series is :-

Options

A. `2.5 : 1`
B. `3.5 : 1`
C. `4.5 : 1`
D. `5.5 : 1`

Explanation

To solve the problem of finding the ratio of the difference in wavelengths of the first and second lines of the Lyman series to the difference in wavelengths of the second and third lines of the same series in a hydrogen-like atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Lyman Series**:
The Lyman series corresponds to transitions where the electron falls to the n=1 energy level. The formula for the wavelength of the emitted light 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, \( n_1 \) is the lower energy level (1 for Lyman series), and \( n_2 \) is the higher energy level (2, 3, 4, ...).

2. **Calculate Wavelengths for the First, Second, and Third Lines**:
- For the **first line** (n=2):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_1 = \frac{4}{3R} \]

- For the **second line** (n=3):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \left( \frac{8}{9} \right) \]
Thus,
\[ \lambda_2 = \frac{9}{8R} \]

- For the **third line** (n=4):
\[ \frac{1}{\lambda_3} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \left( \frac{15}{16} \right) \]
Thus,
\[ \lambda_3 = \frac{16}{15R} \]

3. **Calculate the Differences in Wavelengths**:
- The difference between the first and second lines:
\[ \Delta \lambda_{1,2} = \lambda_1 - \lambda_2 = \frac{4}{3R} - \frac{9}{8R} \]
To combine these fractions, find a common denominator (24R):
\[ \Delta \lambda_{1,2} = \frac{32}{24R} - \frac{27}{24R} = \frac{5}{24R} \]

- The difference between the second and third lines:
\[ \Delta \lambda_{2,3} = \lambda_2 - \lambda_3 = \frac{9}{8R} - \frac{16}{15R} \]
Again, find a common denominator (120R):
\[ \Delta \lambda_{2,3} = \frac{135}{120R} - \frac{128}{120R} = \frac{7}{120R} \]

4. **Calculate the Ratio of the Differences**:
Now, we find the ratio of the two differences:
\[ \text{Ratio} = \frac{\Delta \lambda_{1,2}}{\Delta \lambda_{2,3}} = \frac{\frac{5}{24R}}{\frac{7}{120R}} = \frac{5}{24} \times \frac{120}{7} = \frac{600}{168} = \frac{25}{7} \]

5. **Final Result**:
The ratio of the difference in wavelengths of the first and second lines to the difference in wavelengths of the second and third lines of the Lyman series is:
\[ \frac{25}{7} \]