When a hydrogen atom is excited from ground state to first excited state, then

$\mathrm{KE}=\frac{13.6}{{\mathrm{n}}^2}\mathrm{eV};\mathrm{PE}=\frac{-2\left(13.6\right)}{{\mathrm{n}}^2}\mathrm{eV}$

For hydrogen, Z =1;  $\mathrm{\Delta KE}={\mathrm{KE}}_{\mathrm{f}}-{\mathrm{KE}}_{\mathrm{i}}$

$\mathrm{\Delta KE}=13.6\left[\frac{1}{2^2}-\frac{1}{1^2}\right]=-10.2\mathrm{eV}\left[\mathrm{Decrease}\right]$

$\begin{array}{l}\mathrm{\Delta PE}={\mathrm{PE}}_{\mathrm{f}}-{\mathrm{PE}}_{\mathrm{i}}\\ \mathrm{\Delta PE}=2\left(13.6\right)\left[-\frac{1}{2^2}-\frac{-1}{1^2}\right]=20.4\mathrm{eV}\left[\mathrm{increase}\right]\end{array}$

change in angular momentum, $△\mathrm{L}=\frac{\mathrm{h}}{2\mathrm{\pi }}\left(2-1\right)=\frac{\mathrm{h}}{2\mathrm{\pi }}=1.05\times {10}^{-34}\mathrm{J}-\mathrm{s}$