Question

What is spin angular momentum of $N$ -atom $(Z = 7)$ ? (A) Zero (B) $\sqrt {\dfrac{3}{4}} (\dfrac{h}{{2\pi }})$ (C) $\sqrt {\dfrac{{15}}{4}} (\dfrac{h}{{2\pi }})$ (D) $\dfrac{1}{2}(\dfrac{h}{{2\pi }})$

Answer 1

Hint : Angular momentum is defined as: The property of any rotating object given by moment of inertia times angular velocity.It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object. It is a vector quantity, which implies that here along with magnitude, the direction is also considered.

Complete Step By Step Answer:
The Spin Quantum Number ( ms ) describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude ( $\dfrac{1}{2}$ ) and direction (+ or -). Each orbital can only hold two electrons. One electron will have a +1/2 spin and the other will have a $\dfrac{1}{2}$ spin. Electrons like to fill orbitals before they start to pair up. Therefore the first electron in an orbital will have a spin of $+ \dfrac{1}{2}$ . After all the orbitals are half filled, the electrons start to pair up. This second electron in the orbital will have a spin of $- \dfrac{1}{2}$ . If there are two electrons in the same orbital, it will spin in opposite directions. 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, $N = 1{s^2}2{s^2}2{p^3}$ $s = 3 \times ( \pm \dfrac{1}{2}) = \dfrac{3}{2}$ 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, $s = 3 \times ( \pm \dfrac{1}{2}) = \dfrac{3}{2}$ As we have to calculate spin angular momentum, the formula for the spin angular ${L_s} = \sqrt {s(s + 1)} \dfrac{h}{{2\pi }}$ By putting the value of the total magnetic spin quantum number for the nitrogen atom i.e. in the equation of spin angular momentum we get ${L_s} = \sqrt {\dfrac{3}{2}(\dfrac{3}{2} + 1)} \dfrac{h}{{2\pi }} = \sqrt {\dfrac{{15}}{4}} \dfrac{h}{{2\pi }}$
So, from the above option C is the correct one .

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 $( \pm \dfrac{1}{2})$ 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.