Question

Spin angular momentum for electron is given by:
A) $\sqrt {S\left( {S + 1} \right)} \dfrac{h}{{2\pi }}$
B) $\sqrt {2S\left( {S + 1} \right)} \dfrac{h}{{2\pi }}$
C) $\sqrt {S\left( {S + 2} \right)} \dfrac{h}{{2\pi }}$
D) None

Answer 1

Hint: The spin of an electron is a dimensionless quantity. The spin of an electron is described as the spin quantum number. All the particles that are moving in three directions possess integer spin or half integer spin.

Complete step by step answer:We know that spin of an electron is an intrinsic form of angular momentum carried by elementary particles. The spin of an electron is not associated with rotating internal parts of the elementary particles.
We know that an electron is believed to be a point particle that possesses no internal structure but still the electron possesses spin. A spin quantum number is assigned to the elementary particles that have the same spin angular momentum.
The direction of the spin of the electron can be changed but the electron cannot be made to spin faster or slower.
We know that the spin quantum number $\left( S \right)$ describes the angular momentum of an electron. An electron spins around an axis and possesses both angular momentum and orbital angular momentum. Thus, the spin quantum number has both direction $\left( { + {\text{ or }} - } \right)$ and magnitude $\left( {{\text{1/2}}} \right)$.
According to the definition of spin quantum number,
$S = \dfrac{n}{2}$
Where $S$ is the spin quantum number,
             $n$ is the non-negative integer value.
Thus, the allowed values of spin are $0,\dfrac{1}{2},1,\dfrac{3}{2},......$Thus, the spin is quantised and has only discrete values.
The spin angular momentum of an electron is given as,
$S = h\sqrt {S\left( {S + 1} \right)} $
But $h = \dfrac{h}{{2\pi }}$. Thus,
$S = \dfrac{h}{{2\pi }}\sqrt {S\left( {S + 1} \right)} $
Thus, spin angular momentum for the electron is given by $\sqrt {S\left( {S + 1} \right)} \dfrac{h}{{2\pi }}$.
Thus, the correct option is (A) $\sqrt {S\left( {S + 1} \right)} \dfrac{h}{{2\pi }}$.

Note: The spin angular momentum of an electron can also be represented as $S = \dfrac{h}{{2\pi }}\sqrt {n\left( {n + 2} \right)} $. A particle having $S = \dfrac{h}{2}$ deflects upwards and a particle having $S = - \dfrac{h}{2}$ deflects downwards. The deflection in the upward and the downward direction is the same.