As per Bohr's model the energy (in eV) required to remove electron from ground state of \( \mathrm{Li}^{++} \) (doubly ionised lithium atom) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 122.4
\( \mathrm{Li}^{++} \) has one electron left, so it is a hydrogen-like ion with \( Z = 3 \). In Bohr's model the energy of level \( n \) of such an ion is
\( \displaystyle E_n = -\frac{13.6\,Z^{2}}{n^{2}}\,\mathrm{eV} \)
For the ground state, \( n = 1 \):
\( \displaystyle E_1 = -\frac{13.6 \times 3^{2}}{1^{2}}\,\mathrm{eV} = -122.4\,\mathrm{eV} \)
The energy needed to remove the electron (the ionisation energy) takes it from \( E_1 \) to zero:
\( \displaystyle E_{\mathrm{ion}} = 0 - (-122.4\,\mathrm{eV}) = 122.4\,\mathrm{eV} \)