Which transition in `Li^(2 +)` would have the same wavelength as the `2 rarr 4` transition in `He^(+)` ion ?

Options

A. `4 rarr 2`
B. `2 rarr 4`
C. `3 rarr 6`
D. `6 rarr 2`

Explanation

To solve the problem of determining which transition in \( \text{Li}^{2+} \) would have the same wavelength as the \( 2 \rightarrow 4 \) transition in \( \text{He}^{+} \), we can follow these steps:

### Step 1: Understand the Wavelength Formula
The wavelength \( \lambda \) of the emitted or absorbed light during a transition in a hydrogen-like atom can be calculated using the formula:
\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]
where:
- \( R_H \) is the Rydberg constant,
- \( Z \) is the atomic number,
- \( n_i \) is the initial energy level,
- \( n_f \) is the final energy level.

### Step 2: Calculate for \( \text{He}^{+} \)
For the \( \text{He}^{+} \) ion, which has \( Z = 2 \), we need to calculate the wavelength for the transition \( 2 \rightarrow 4 \).

Using the formula:
\[ \frac{1}{\lambda_{He}} = R_H \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Calculating this:
\[ \frac{1}{\lambda_{He}} = R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = R_H \cdot 4 \left( \frac{4 - 1}{16} \right) = R_H \cdot 4 \cdot \frac{3}{16} = \frac{3 R_H}{4} \]

### Step 3: Set Up for \( \text{Li}^{2+} \)
Now, for \( \text{Li}^{2+} \), which has \( Z = 3 \), we need to find the transition that gives the same wavelength.

Using the same formula:
\[ \frac{1}{\lambda_{Li}} = R_H \cdot 3^2 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda_{Li}} = R_H \cdot 9 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]

### Step 4: Equate the Two Wavelengths
Since we want the wavelengths to be equal:
\[ \frac{3 R_H}{4} = R_H \cdot 9 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]
Cancelling \( R_H \) from both sides:
\[ \frac{3}{4} = 9 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]
Dividing both sides by 9:
\[ \frac{1}{12} = \frac{1}{n_i^2} - \frac{1}{n_f^2} \]

### Step 5: Find Suitable \( n_i \) and \( n_f \)
We need to find integers \( n_i \) and \( n_f \) such that:
\[ \frac{1}{n_i^2} - \frac{1}{n_f^2} = \frac{1}{12} \]

Testing possible transitions:
- For \( n_i = 3 \) and \( n_f = 6 \):
\[ \frac{1}{3^2} - \frac{1}{6^2} = \frac{1}{9} - \frac{1}{36} = \frac{4 - 1}{36} = \frac{3}{36} = \frac{1}{12} \]

### Conclusion
The transition \( 3 \rightarrow 6 \) in \( \text{Li}^{2+} \) has the same wavelength as the \( 2 \rightarrow 4 \) transition in \( \text{He}^{+} \).

Thus, the answer is:
**3 to 6 (Option C)**