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