Energy required for the electron excitation in $\mathrm{Li}^{++}$from the first to the third Bohr orbit is :

Question

Aicharya · Accepted Answer

$E_n=-13.6 \frac{Z^2}{n^2}$ $E_{\mathrm{L}}{ }^{* *}=-13.6 	imes \frac{9}{1}=-122.4 \mathrm{eV}$ $E_{\mathrm{L}}{ }^{* *}=-13.6 	imes \frac{9}{9}=-13.6 \mathrm{eV}$ $\Delta E=-13.6-(-122.4)$ $=108.8 \mathrm{eV}$

Energy required for the electron excitation in $\mathrm{Li}^{++}$from the first to the third Bohr orbit is :

Solution