MCQ
As per Bohr model, the minimum energy (in $eV$) required to remove an electron from the ground state of doubly ionized $Li$ atom $(Z = 3)$ is
- A$1.51$
- B$13.6$
- C$40.8$
- ✓$122.4$
✓
Answer
Correct option: D.
$122.4$
d
(d) $E = - {Z^2} \times 13.6\;eV$$ = - 9 \times 13.6\;eV = - 122.4\;eV$
So ionisation energy $= + 122.4 \,eV.$
(d) $E = - {Z^2} \times 13.6\;eV$$ = - 9 \times 13.6\;eV = - 122.4\;eV$
So ionisation energy $= + 122.4 \,eV.$
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