Which electeonic transition in `Li^(+2)` will emit photon of same energy as that emitted due to transition from ` 5^(th) ` excited state to `1^(st)` excited State in `He^(+)`.

Options

A. `6to2`
B. `3to1`
C. `9to3`
D. `2to6`

Explanation

To solve the problem, we need to find the electronic transition in \( \text{Li}^{+2} \) that emits a photon of the same energy as that emitted during the transition from the 5th excited state to the 1st excited state in \( \text{He}^{+} \).

### Step-by-Step Solution:

1. **Identify the Transition in \( \text{He}^{+} \)**:
- The 5th excited state corresponds to \( n = 6 \) (since the ground state is \( n = 1 \)).
- The 1st excited state corresponds to \( n = 2 \).
- Therefore, the transition in \( \text{He}^{+} \) is from \( n = 6 \) to \( n = 2 \).

2. **Calculate the Energy of the Transition in \( \text{He}^{+} \)**:
- The energy of the transition can be expressed using the formula:
\[ E = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For \( \text{He}^{+} \) (where \( Z = 2 \)):
\[ E_{\text{He}^{+}} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \]
- Simplifying this gives:
\[ E_{\text{He}^{+}} = 4R \left( \frac{1}{4} - \frac{1}{36} \right) = 4R \left( \frac{9 - 1}{36} \right) = 4R \cdot \frac{8}{36} = \frac{32R}{36} = \frac{8R}{9} \]

3. **Set Up the Energy Equation for \( \text{Li}^{+2} \)**:
- For \( \text{Li}^{+2} \) (where \( Z = 3 \)):
\[ E_{\text{Li}^{+2}} = R \cdot 3^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 9R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Equate the Energies**:
- Set the energies equal to each other:
\[ 9R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{8R}{9} \]
- Dividing both sides by \( R \) gives:
\[ 9 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{8}{9} \]
- Rearranging gives:
\[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{8}{81} \]

5. **Choose Values for \( n_1 \) and \( n_2 \)**:
- We need to find integers \( n_1 \) and \( n_2 \) such that:
\[ n_2 > n_1 \]
- Let's try \( n_1 = 3 \) and \( n_2 = 9 \):
\[ \frac{1}{3^2} - \frac{1}{9^2} = \frac{1}{9} - \frac{1}{81} = \frac{9 - 1}{81} = \frac{8}{81} \]
- This satisfies our equation.

6. **Conclusion**:
- The electronic transition in \( \text{Li}^{+2} \) that emits a photon of the same energy as that emitted due to the transition from the 5th excited state to the 1st excited state in \( \text{He}^{+} \) is from \( n = 9 \) to \( n = 3 \).

### Final Answer:
The transition is from the 9th state to the 3rd state in \( \text{Li}^{+2} \).