What will be the angular momentum of a electron, if energy of this electron in H-atom is 1.5eV (in J-sec)

3.15 × 10^-34

Energy of electron in H atom $E_n= \frac{-13.6}{n^2} ~eV$

$\Rightarrow-1.5=\frac{-13.6}{n^2} \Rightarrow n^2=\frac{13.6}{1.5}=3$

Now angular momentum

$p=n \frac{h}{2 \pi}=\frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.14}=3.15 \times 10^{-34} J \times sec$