The ratio of the wavelengths for `2 rarr 1` transition in `Li^(++), He^(+)` and `H` is

Options

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

Explanation

To find the ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\), we can use the formula for the wavelength of the emitted radiation during a transition in a hydrogen-like atom:

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

Where:
- \(R\) is the Rydberg constant,
- \(Z\) is the atomic number,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels, respectively.

### Step 1: Identify the values of \(Z\) for each ion
- For \(Li^{++}\), \(Z = 3\)
- For \(He^{+}\), \(Z = 2\)
- For \(H\), \(Z = 1\)

### Step 2: Set the values of \(n_1\) and \(n_2\)
In this case, \(n_1 = 1\) and \(n_2 = 2\).

### Step 3: Calculate \(\frac{1}{\lambda}\) for each ion
Using the formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating for each ion:

1. **For \(Li^{++}\)**:
\[ \frac{1}{\lambda_{Li}} = R(3^2) \left( \frac{1}{1} - \frac{1}{4} \right) = R(9) \left( \frac{3}{4} \right) = \frac{27R}{4} \]

2. **For \(He^{+}\)**:
\[ \frac{1}{\lambda_{He}} = R(2^2) \left( \frac{1}{1} - \frac{1}{4} \right) = R(4) \left( \frac{3}{4} \right) = 3R \]

3. **For \(H\)**:
\[ \frac{1}{\lambda_{H}} = R(1^2) \left( \frac{1}{1} - \frac{1}{4} \right) = R(1) \left( \frac{3}{4} \right) = \frac{3R}{4} \]

### Step 4: Find the wavelengths \(\lambda\)
The wavelengths can be expressed as:
\[ \lambda_{Li} = \frac{4}{27R}, \quad \lambda_{He} = \frac{1}{3R}, \quad \lambda_{H} = \frac{4}{3R} \]

### Step 5: Calculate the ratio of wavelengths
Now we find the ratio of the wavelengths:
\[ \frac{\lambda_{Li}}{\lambda_{He}} = \frac{\frac{4}{27R}}{\frac{1}{3R}} = \frac{4}{27} \cdot 3 = \frac{12}{27} = \frac{4}{9} \]

\[ \frac{\lambda_{He}}{\lambda_{H}} = \frac{\frac{1}{3R}}{\frac{4}{3R}} = \frac{1}{4} \]

\[ \frac{\lambda_{Li}}{\lambda_{H}} = \frac{\frac{4}{27R}}{\frac{4}{3R}} = \frac{4}{27} \cdot \frac{3}{4} = \frac{3}{27} = \frac{1}{9} \]

### Final Ratio
Thus, the final ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\) is:
\[ \lambda_{Li} : \lambda_{He} : \lambda_{H} = \frac{4}{9} : \frac{1}{4} : \frac{1}{9} \]