Let the value of angular momentum of orbit, orbital angular momentum and spin angular momentum for the outermost electron in sulphur atom  $(\mathrm{Z}=16)\text{be }{n}_1\frac{h}{2\pi },\sqrt{n_2}\frac{h}{2\pi } \text{and }\sqrt{n_3}\frac{h}{4\pi }$ . Find the value of $(n_1+n_2+n_3)$.

For the last electron in S-atom ; $l=1 \text{and }s=+\frac{1}{2}, n=3$

Angular momentum of orbit  $=\frac{n_{1}h}{2\pi },$  $\therefore n_1=3$

Orbital angular momentum   $=\sqrt{l(l+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi }=\sqrt{n_2}\frac{h}{2\pi }$   $\therefore n_2=2$

Spin angular momentum $=\sqrt{s(s+1)}\frac{h}{2\pi }=\sqrt{3}\frac{h}{4\pi }=\sqrt{n_3}\frac{h}{4\pi }$   $\therefore n_3=3$

$\therefore n_1+n_2+n_3=3+2+3=8$