The $I{E}_1$ of $Li$ is $5.4eV$ and $I{E}_1$ of $H$ is $13.6eV$. Calculate the charge acting on the outermost electron of $Li$ atom.

Solution: For $Li$. electronic configureation si $1s^2,2s^2,$ so given ionisation energy value is for $\left(n=2\right)$.
We know that : energy of $n^{th}$ level $E_{eff}^2\times E_1$
$\left[\begin{array}{c}\text{Here}{E}_1=\text{Energy of first orbit of}\text{H-atom}\\ Z_{eff}=\text{Effective nuclear cahrge}\end{array}\right]$
or $Z_{eff}=n\sqrt{\frac{D_2}{E_1}}$
(Given: $E_1=-13.6eV,E_2=-5.39eV$ and $\left(n=2\right)$
$Z_{eff}=2\times \frac{\sqrt{\left(5.39\right)}}{13.6}=1..26$
Thus effective nuclear charge si $1.26eV1$ because $2s$.