As per Bohr’s model, what energy (in eV) must be supplied to remove an electron from the ground state of the doubly ionized Li atom
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 122.4\ \mathrm{eV} \)
Doubly ionised lithium, \( \mathrm{Li^{2+}} \), has one electron around a nucleus of charge \( Z = 3 \). It is hydrogen-like, so Bohr’s energy formula applies:
\( \displaystyle E_{n} = -\frac{13.6\,Z^{2}}{n^{2}}\ \mathrm{eV} \)
\( E_{1} = -13.6\times3^{2}\,\mathrm{eV} = -122.4\,\mathrm{eV} \)
To remove the electron from the ground state, it must be raised to \( E = 0 \), so the ionisation energy is:
\( 0 - (-122.4\,\mathrm{eV}) = 122.4\,\mathrm{eV} \)