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

Options

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

Answer

To solve the problem of finding the ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\), we can use the Rydberg formula for the wavelengths of spectral lines in hydrogen-like atoms.

### Step-by-Step Solution:

1. **Understanding the Rydberg Formula**:
The Rydberg formula for the wavelength (\(\lambda\)) of light emitted during an electronic transition in a hydrogen-like atom 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,
- \(n_1\) and \(n_2\) are the principal quantum numbers of the lower and upper energy levels respectively.

2. **Identifying the Transition**:
For the transition \(2 \rightarrow 1\):
- \(n_1 = 1\)
- \(n_2 = 2\)

3. **Calculating the Wavelengths**:
For each atom, we will calculate \(\frac{1}{\lambda}\):
- For Hydrogen (\(H\), \(Z = 1\)):
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
- For Helium (\(He^{+}\), \(Z = 2\)):
\[ \frac{1}{\lambda_{He}} = R \cdot 2^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 4 \left( 1 - \frac{1}{4} \right) = R \cdot 4 \cdot \frac{3}{4} = 3R \]
- For Lithium (\(Li^{++}\), \(Z = 3\)):
\[ \frac{1}{\lambda_{Li}} = R \cdot 3^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 9 \left( 1 - \frac{1}{4} \right) = R \cdot 9 \cdot \frac{3}{4} = \frac{27R}{4} \]

4. **Finding the Wavelengths**:
Now we can express the wavelengths:
- \(\lambda_H = \frac{4}{3R}\)
- \(\lambda_{He} = \frac{1}{3R}\)
- \(\lambda_{Li} = \frac{4}{27R}\)

5. **Calculating the Ratios**:
We need to find the ratio \(\lambda_{Li} : \lambda_{He} : \lambda_H\):
\[ \lambda_{Li} : \lambda_{He} : \lambda_H = \frac{4}{27R} : \frac{4}{3R} : \frac{4}{3R} \]
To simplify, we can multiply through by \(27R\):
\[ 4 : 36 : 36 \]
This simplifies to:
\[ 1 : 9 : 9 \]

6. **Final Ratio**:
The final ratio of the wavelengths for the \(2 \rightarrow 1\) transition in \(Li^{++}\), \(He^{+}\), and \(H\) is:
\[ \lambda_{Li} : \lambda_{He} : \lambda_H = 1 : 9 : 9 \]