What will be the energy required to remove the electron from the first excited state of \( \mathrm{Li^{++}} \), if the ground state energy of the electron in a hydrogen atom is \( -13.6\,\mathrm{eV} \)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 30.6 eV
\( \mathrm{Li^{++}} \) has a single electron, so it is a hydrogen-like ion and the Bohr model applies with \( Z = 3 \).
\( \displaystyle E_n = -\frac{13.6\,Z^{2}}{n^{2}}\,\mathrm{eV} \)
The first excited state is \( n = 2 \).
\( \displaystyle E_2 = -\frac{13.6 \times 3^{2}}{2^{2}} = -\frac{13.6 \times 9}{4} = -30.6\,\mathrm{eV} \)
The energy needed to remove the electron from this state takes it to \( E = 0 \), so it is \( 0 - (-30.6) = 30.6\,\mathrm{eV} \).