Which of the following is the correct value for the first ionization energy of doubly ionized lithium (Li^{2+}), where Z = 3? Use the formula E = -13.6 ext{ eV} imes Z^2 / n^2 for hydrogen-like ions, and recall that the first ionization energy is the energy required to remove the only electron from the n = 1 state.

• A. 122.4 eV
• B. 13.6 eV
• C. 40.8 eV
• D. 61.2 eV

Answer: (A)

Verified step by step guidance

1. Identify the given information: the ion is doubly ionized lithium (Li^{2+}), which means it has only one electron left, making it a hydrogen-like ion with atomic number Z = 3, and the electron is in the n = 1 energy level.
2. Recall the formula for the energy of a hydrogen-like ion: \[E = -13.6 \times \frac{Z^2}{n^2} \text{ eV}\] where \(Z\) is the atomic number and \(n\) is the principal quantum number of the electron's orbit.
3. Substitute the known values into the formula: \[E = -13.6 \times \frac{3^2}{1^2} = -13.6 \times 9\] This gives the energy of the electron in the n = 1 state for Li^{2+}.
4. Understand that the first ionization energy is the energy required to remove the electron from this state, which is the positive value of the energy magnitude (since the electron is bound with negative energy).
5. Calculate the magnitude of the energy (without the negative sign) to find the first ionization energy, which corresponds to the energy needed to ionize Li^{2+} from the n = 1 state.