Physics - Modern Physics Question with Solution | TestHub

Energy of an electron in n^th orbit of an ion with atomic number $z$ is, $E=-13.6\times \frac{z^2}{n^2}eV$

for $1^{st}$ excited state of $L{i}^{++},\text{z}=3,n=2$

$E=-13.6\times \frac{3^2}{n^2}=-13.6\times \frac{9}{4}eV$

$=-30.6eV$

$\therefore$ Required energy $=30.6eV$