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

### Step-by-Step Solution:

1. **Understand the Formula for Wavelength**:
The wavelength (\( \lambda \)) of the emitted or absorbed light during a transition in a hydrogen-like atom can be calculated using the formula:
\[ \frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states, respectively.

2. **Calculate Wavelength for \( He^+ \) Transition**:
For the \( He^+ \) ion, \( Z = 2 \) and the transition is from \( n_1 = 2 \) to \( n_2 = 4 \):
\[ \frac{1}{\lambda_{He}} = R_H \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Simplifying this:
\[ \frac{1}{\lambda_{He}} = R_H \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \cdot 4 \left( \frac{4 - 1}{16} \right) = R_H \cdot 4 \cdot \frac{3}{16} = \frac{3R_H}{4} \]
Thus,
\[ \lambda_{He} = \frac{4}{3R_H} \]

3. **Set Up the Equation for \( Li^{2+} \)**:
For \( Li^{2+} \), \( Z = 3 \). We need to find the transition \( n_1 \) to \( n_2 \) such that:
\[ \frac{1}{\lambda_{Li}} = R_H \cdot 3^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Setting \( \lambda_{Li} = \lambda_{He} \):
\[ \frac{1}{\lambda_{Li}} = \frac{3R_H}{4} \]
Therefore,
\[ R_H \cdot 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R_H}{4} \]
Dividing both sides by \( R_H \):
\[ 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{4} \]
Simplifying gives:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{12} \]

4. **Finding Suitable \( n_1 \) and \( n_2 \)**:
We need to find integers \( n_1 \) and \( n_2 \) such that:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{12} \]
Testing possible transitions:
- For \( n_1 = 3 \) and \( n_2 = 6 \):
\[ \frac{1}{3^2} - \frac{1}{6^2} = \frac{1}{9} - \frac{1}{36} = \frac{4 - 1}{36} = \frac{3}{36} = \frac{1}{12} \]
This satisfies our equation.

5. **Conclusion**:
The transition in \( Li^{2+} \) that has the same wavelength as the \( 2 \rightarrow 4 \) transition in \( He^+ \) is from \( n_1 = 3 \) to \( n_2 = 6 \).

### Final Answer:
The transition in \( Li^{2+} \) that has the same wavelength as the \( 2 \rightarrow 4 \) transition in \( He^+ \) is \( 3 \rightarrow 6 \).

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