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, we need to find 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 for a hydrogen-like atom.

### Step-by-Step Solution:

1. **Identify the Lyman Series Formula**:
The formula for the wavelengths in the Lyman series is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)), and \( n_1 \) and \( n_2 \) are the principal quantum numbers.

2. **Calculate Wavelengths for the First Three Lines**:
- **First Line (n2 = 2, n1 = 1)**:
\[ \frac{1}{\lambda_1} = R \left( 1 - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Thus,
\[ \lambda_1 = \frac{4}{3R} \]

- **Second Line (n2 = 3, n1 = 1)**:
\[ \frac{1}{\lambda_2} = R \left( 1 - \frac{1}{3^2} \right) = R \left( 1 - \frac{1}{9} \right) = R \cdot \frac{8}{9} \]
Thus,
\[ \lambda_2 = \frac{9}{8R} \]

- **Third Line (n2 = 4, n1 = 1)**:
\[ \frac{1}{\lambda_3} = R \left( 1 - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = R \cdot \frac{15}{16} \]
Thus,
\[ \lambda_3 = \frac{16}{15R} \]

3. **Calculate Differences in Wavelengths**:
- **Difference between First and Second Lines**:
\[ \Delta \lambda_{1,2} = \lambda_1 - \lambda_2 = \frac{4}{3R} - \frac{9}{8R} = \frac{32 - 27}{24R} = \frac{5}{24R} \]

- **Difference between Second and Third Lines**:
\[ \Delta \lambda_{2,3} = \lambda_2 - \lambda_3 = \frac{9}{8R} - \frac{16}{15R} = \frac{135 - 128}{120R} = \frac{7}{120R} \]

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

5. **Final Answer**:
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:
\[ \frac{25}{7} \quad \text{or approximately } 3.57:1 \]