Question

Spin angular momentum of nitrogen atom in ground state is:

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

Answer 1

Answer: (D)

$\sqrt{\frac{15}{4}}\left(\frac{h}{2\pi }\right)$

N(Z=7) $1s^{2}2s^{2}2p^3$

Total magnetic spin quantum numbers = $3(\pm \frac{1}{2})$ due to three unpaired electrons

Spin angular momentum is given by $\sqrt{s(s+1)\frac{h}{2\pi }}$

Now $|s|=\frac{3}{2}$

Required momentum = $\sqrt{\frac{3}{2}(\frac{3}{2}+1)\frac{h}{2\pi }}=\sqrt{\frac{15}{4}}\frac{h}{2\pi }$