Question

The magnitude of the spin angular momentum of an electron is given by:
$(a){\text{ }}S = \sqrt {s(s + 1)} \dfrac{h}{{2\pi }}$
$(b){\text{ }}S = s\dfrac{h}{{2\pi }}$
$(c){\text{ }}S = \dfrac{{\sqrt 3 }}{2} \times \dfrac{h}{{2\pi }}$
$(d){\text{ }}S = \pm \dfrac{1}{2} \times \dfrac{h}{{2\pi }}$

Answer 1

Hint: The spin is given as a dimensionless spin quantum number by dividing the spin angular momentum by the reduced Planck's constant. All particles that move in all the 3 directions have either the integer spin or the half integer spin.

Complete answer:
The spin is an intrinsic form of angular momentum carried by elementary particles and is not associated with rotating internal parts of the elementary particles. An electron is believed to be a point particle and possessing no internal structure but still then the electron has spin. A spin quantum number is assigned to the elementary particles of the given kind that have the same spin angular momentum. Although the direction of the spin of the particle can be changed, it cannot be made to spin faster or slower.
According to the conventional definition of the spin quantum number $S = \dfrac{n}{2}$ , where $n$ can be any non-negative integer value. So, we get the allowed values of spin as $0,\dfrac{1}{2},1,\dfrac{3}{2},......$etc. Thus, the spin is quantized and can take only discrete values. The elementary values for which $S = \dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},.....$are called Fermions whereas the elementary particles for which $S = 0,1,2,.....$are called Bosons.
The spin angular momentum S of any physical system is given as
$S = \hbar \sqrt {s(s + 1)} = \dfrac{h}{{4\pi }}\sqrt {n(n + 2)} $
Where $h$ is the Planck's constant and $\hbar $ is the reduced Planck's constant such that $\hbar = \dfrac{h}{{2\pi }}$.

Hence, the correct solution is the option$(a)$.

Note: The spin angular momentum can also be represented as $S = \dfrac{h}{{2\pi }}\sqrt {n(n + 2)} $. A particle with $S = \dfrac{\hbar }{2}$is deflected upwards whereas a particle with $S = - \dfrac{\hbar }{2}$is deflected downwards. Also, the amount of deflection in the upward as well as in the downward direction is the same.