Question

If the bonding energy of the electron in a hydrogen atom is 13.6 eV, then the energy required to remove an electron from the first excited state of Li$^{2+}$ is

• A. 122.4 eV
• B. 3.4 eV
• C. 13.6 eV
• D. 30.6 eV

Hint:

For hydrogen-like atoms, the energy required to remove an electron follows $E_n = \frac{13.6 Z^2}{n^2}$.

Answer 1

The Correct Option is D

Energy levels for hydrogen-like atoms are given by: $$E_n = \frac{13.6 Z^2}{n^2} \text{ eV}$$ For Li$^{2+}$ (Z = 3), the first excited state corresponds to $n = 2$: $$E_2 = \frac{13.6 \times 9}{4} = 30.6 \text{ eV}$$ Thus, the correct answer is 30.6 eV.