Question

Spin angular momentum of a nitrogen atom in the ground state is:
A) $\dfrac{1}{2}\dfrac{h}{{2\pi }}$
B) Zero
C) $\sqrt {\dfrac{3}{4}} \left( {\dfrac{h}{{2\pi }}} \right)$
D) $\sqrt {\dfrac{{15}}{4}} \left( {\dfrac{h}{{2\pi }}} \right)$

Answer 1

Hint: In a ground state, the nitrogen atom has an atomic number seven which can be used to configure the electronic configuration of the nitrogen. By using this data one can calculate the total magnetic spin quantum numbers of nitrogen and can find out spin angular momentum.

 Complete step by step answer:
1) First of all we have to understand that the nitrogen atom is in the ground state that means it has the normal electrons in its orbital. Now let's write the electronic configuration of the nitrogen atom which has the atomic number seven,
$N = 1{s^2}2{s^2}2{p^3}$ (as in the ground state)
2) The $2p$ orbital can have a maximum of six electrons but there are only three present in the $2p$ orbital of the nitrogen atom. Hence, the nitrogen atom has three unpaired electrons in its $2p$ orbital. So, the total magnetic spin quantum number for the nitrogen atom which has three unpaired electrons will be, $s = 3 \times \left( { \pm \dfrac{1}{2}} \right) = \dfrac{3}{2}$
3) As we have to calculate spin angular momentum, the formula for the spin angular momentum is ${L_s} = \sqrt {s\left( {s + 1} \right)\dfrac{h}{{2\pi }}} $
4) By putting the value of the total magnetic spin quantum number for the nitrogen atom i.e. $s$ in the equation of spin angular momentum we get,
${L_s} = \sqrt {\dfrac{3}{2}\left( {\dfrac{3}{2} + 1} \right)\dfrac{h}{{2\pi }}} = \sqrt {\dfrac{{15}}{4}} \left( {\dfrac{h}{{2\pi }}} \right)$
5) The obtained value of spin angular momentum is $\sqrt {\dfrac{{15}}{4}} \left( {\dfrac{h}{{2\pi }}} \right)$ which is given in option D. Therefore, the option D is the correct choice of answer.

So, the correct answer is Option D.

Note:
While the calculation of the total magnetic spin quantum number for an element the sum of all the unpaired electrons is done where each electron stands for $\left( { \pm \dfrac{1}{2}} \right)$ in which the sign $ \pm $ shows the direction of the spin of electrons. Spin of an element is a fundamental property. Angular momentum of an element is a vector quantity.