Question

What is the spin angular momentum of N-atom (Z=7)?

$EC=1s^{2}2s^{2}2p^3\left(GS\right)$

• A. Zero
• B. $\sqrt{\frac{3}{4}}\left(\frac{h}{2\pi }\right)$
• C. $\sqrt{\frac{15}{4}}\left(\frac{h}{2\pi }\right)$
• D. $\frac{1}{2}\left(\frac{h}{2\pi }\right)$

Answer 1

The correct option is C $\sqrt{\frac{15}{4}}\left(\frac{h}{2\pi }\right)$

Electronic configuration of N-atom (Z=7): $EC=1s^{2}2s^{2}2p^3\left(GS\right)$, so it has 3 unpaired electrons and sum of spin quantum number is (3/2).

A spin angular momentum, $L_s=\sqrt{s(s+1)}\frac{h}{2\pi }$, where S = sum of spin Q.N. of all the electrons.

Spin angular momentum $=\sqrt{\frac{15}{4}}\left(\frac{h}{2\pi }\right)$