Question

Let the value of angular momentum, orbital angular momentum and spin angular momentum for the last electron in sulphur atom in Bohr orbit be $n_1\frac{h}{4\pi },\sqrt{n_2}\frac{h}{4\pi }$ and $\sqrt{n_3}\frac{h}{4\pi }$, respectively. The value of $(n_1+n_2+n_3)\times 2$ in Bohr orbit is:

(If the multiple value is 56, then write answer is the last digit as 6)

Answer 1

For the last electron in S-atom: $l=1,n=3.$
Angular momentum in Bohr orbit $=\frac{n_{1}h}{2\pi }=\frac{3h}{2\pi }=\frac{6h}{4\pi }$
$\therefore n_1=6$
Orbital angular momentum $=\sqrt{l(l+1)}\frac{h}{2\pi }=\sqrt{2}\frac{h}{2\pi }=$
$\sqrt{8}\frac{h}{4\pi }=\sqrt{n_2}\frac{h}{4\pi }$ $\therefore n_2=8$
Spin angular momentum $=\sqrt{s(s+1)}\frac{h}{2\pi }=\sqrt{3}\frac{h}{4\pi }=$$\sqrt{n_3}\frac{h}{4{\pi }_3}\Rightarrow n_3=3.$
$\therefore n_1+n_2+n_3=6+8+3=17.$

Answer $17\times 2=34\equiv 4$.