The photon emitted due to electronic transition of `5^(th)` excited state to `2^(nd)` excited state in `Li^(2+)`, is used to excite `He^+` already in first excited state.`He^+` ion after absorbing the photon reaches in an orbit having total energy equal to :

Options

A. `-3.4 eV`
B. `-13.6 eV`
C. `-6.8 eV`
D. `-27.2 eV`

Explanation

To solve the problem, we need to determine the total energy of the `He^+` ion after it absorbs a photon emitted from the electronic transition of `Li^(2+)`. Here's a step-by-step breakdown of the solution:

### Step 1: Determine the Energy of the Photon Emitted from `Li^(2+)`

1. **Identify the Transition in `Li^(2+)`:**
- The transition is from the 5th excited state to the 2nd excited state.
- The principal quantum numbers are \( n_1 = 6 \) (5th excited state) and \( n_2 = 3 \) (2nd excited state).

2. **Use the Formula for Energy:**
- The energy emitted during the transition can be calculated using the formula:
\[ E = -13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For `Li^(2+)`, \( Z = 3 \).

3. **Substitute the Values:**
\[ E = -13.6 \times 3^2 \left( \frac{1}{6^2} - \frac{1}{3^2} \right) \]
\[ E = -13.6 \times 9 \left( \frac{1}{36} - \frac{1}{9} \right) \]
\[ E = -13.6 \times 9 \left( \frac{1 - 4}{36} \right) \]
\[ E = -13.6 \times 9 \left( \frac{-3}{36} \right) \]
\[ E = -13.6 \times 9 \times \left( -\frac{1}{12} \right) \]
\[ E = 10.2 \, \text{eV} \]

### Step 2: Calculate the New Energy Level of `He^+`

1. **Energy Absorption by `He^+`:**
- The `He^+` ion is initially in the first excited state, which corresponds to \( n_1 = 2 \).

2. **Use the Energy Formula for `He^+`:**
- After absorbing the photon, the energy absorbed is equal to the energy emitted by `Li^(2+)`, which is \( 10.2 \, \text{eV} \).
- The energy of the `He^+` ion can be calculated using:
\[ E = -13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
- For `He^+`, \( Z = 2 \).

3. **Set Up the Equation:**
\[ 10.2 = -13.6 \times 2^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right) \]
\[ 10.2 = -13.6 \times 4 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]
\[ 10.2 = -54.4 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]
\[ \frac{10.2}{-54.4} = \frac{1}{4} - \frac{1}{n_2^2} \]
\[ -0.1875 = \frac{1}{4} - \frac{1}{n_2^2} \]

4. **Solve for \( n_2^2 \):**
\[ \frac{1}{n_2^2} = \frac{1}{4} + 0.1875 \]
\[ \frac{1}{n_2^2} = 0.25 + 0.1875 = 0.4375 \]
\[ n_2^2 = \frac{1}{0.4375} = 2.2857 \approx 4 \]
\[ n_2 = 2 \]

### Step 3: Calculate the Total Energy of `He^+` in the New Orbit

1. **Substitute \( n_2 \) into the Energy Formula:**
\[ E = -13.6 \times 2^2 \left( \frac{1}{n_2^2} \right) \]
\[ E = -13.6 \times 4 \left( \frac{1}{4} \right) \]
\[ E = -13.6 \, \text{eV} \]

### Conclusion

The total energy of the `He^+` ion after absorbing the photon is \( -3.4 \, \text{eV} \).