To find the energy associated with the de-excitation of an electron from \( n=6 \) to \( n=2 \) in the \( He^+ \) ion, we can use the formula for the energy levels of hydrogen-like ions:
\[
E = -\frac{13.6 Z^2}{n^2}
\]
where:
- \( E \) is the energy in electron volts (eV),
- \( Z \) is the atomic number (for helium, \( Z = 2 \)),
- \( n \) is the principal quantum number.
### Step 1: Calculate the energy for \( n=2 \)
Using the formula:
\[
E_2 = -\frac{13.6 \times 2^2}{2^2} = -\frac{13.6 \times 4}{4} = -13.6 \, \text{eV}
\]
### Step 2: Calculate the energy for \( n=6 \)
Now, calculate the energy for \( n=6 \):
\[
E_6 = -\frac{13.6 \times 2^2}{6^2} = -\frac{13.6 \times 4}{36} = -\frac{54.4}{36} \approx -1.5111 \, \text{eV}
\]
### Step 3: Calculate the energy difference (\( \Delta E \))
The energy associated with the transition from \( n=6 \) to \( n=2 \) is given by:
\[
\Delta E = E_2 - E_6
\]
Substituting the values we calculated:
\[
\Delta E = (-13.6) - (-1.5111) = -13.6 + 1.5111 = -12.0889 \, \text{eV}
\]
### Step 4: Convert the energy to kilojoules per mole
To convert the energy from eV to kJ/mol, we use the conversion factor \( 1 \, \text{eV} = 96.485 \, \text{kJ/mol} \):
\[
\Delta E (\text{kJ/mol}) = -12.0889 \, \text{eV} \times 96.485 \, \text{kJ/mol} \approx -1163.5 \, \text{kJ/mol}
\]
### Step 5: Report the final answer
Since we are interested in the magnitude of energy associated with the de-excitation, we take the absolute value:
\[
\Delta E \approx 1163.5 \, \text{kJ/mol}
\]
Thus, rounding to three significant figures, the answer is approximately:
\[
\Delta E \approx 1.16 \times 10^3 \, \text{kJ/mol}
\]
### Final Answer
The correct option is (c) \( 1.16 \times 10^3 \).
