The wavelength of the first Lyman lines of hydrogen `He^(+)` and `Li^(2+)` ions and `lamda_(1),lamda_(2)& lamda_(3)` the ratio of these wavelengths is

Options

A. `1:4:9`
B. `9:4:1`
C. `36:9:4`
D. `6:3:2`

Explanation

To solve the problem of finding the ratios of the wavelengths of the first Lyman lines of hydrogen (H), helium ion (He⁺), and lithium ion (Li²⁺), we can follow these steps:

### Step 1: Understand the Lyman Series
The Lyman series corresponds to electronic transitions in hydrogen-like atoms where an electron falls from a higher energy level (n=2, 3, ...) to the first energy level (n=1). The formula for the wavelength (λ) of the emitted light during these transitions is given by:

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_1 \) is the lower energy level (1 for Lyman series),
- \( n_2 \) is the higher energy level (2 for the first line of Lyman series).

### Step 2: Calculate for Each Ion
1. **For Hydrogen (H, Z=1)**:
\[ \frac{1}{\lambda_1} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Therefore,
\[ \lambda_1 = \frac{4}{3R} \]

2. **For Helium Ion (He⁺, Z=2)**:
\[ \frac{1}{\lambda_2} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 4 \left( 1 - \frac{1}{4} \right) = R \cdot 4 \cdot \frac{3}{4} = 3R \]
Therefore,
\[ \lambda_2 = \frac{1}{3R} \]

3. **For Lithium Ion (Li²⁺, Z=3)**:
\[ \frac{1}{\lambda_3} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 9 \left( 1 - \frac{1}{4} \right) = R \cdot 9 \cdot \frac{3}{4} = \frac{27R}{4} \]
Therefore,
\[ \lambda_3 = \frac{4}{27R} \]

### Step 3: Find the Ratios
Now we have:
- \( \lambda_1 = \frac{4}{3R} \)
- \( \lambda_2 = \frac{1}{3R} \)
- \( \lambda_3 = \frac{4}{27R} \)

To find the ratios \( \lambda_1 : \lambda_2 : \lambda_3 \):
- Multiply each wavelength by \( 27R \) to eliminate \( R \):
- \( \lambda_1 \cdot 27R = \frac{4 \cdot 27}{3} = 36 \)
- \( \lambda_2 \cdot 27R = \frac{1 \cdot 27}{3} = 9 \)
- \( \lambda_3 \cdot 27R = \frac{4 \cdot 27}{27} = 4 \)

Thus, the ratio is:
\[ \lambda_1 : \lambda_2 : \lambda_3 = 36 : 9 : 4 \]

### Final Answer
The ratio of the wavelengths of the first Lyman lines of hydrogen, helium ion, and lithium ion is:
\[ \boxed{36 : 9 : 4} \]