Question

The angular momentum of an electron due to its spin is given as:
A.$\sqrt{s(s+1)}\dfrac{h}{2\pi }$
B.$\sqrt{s(s-1)}\dfrac{h}{2\pi }$
C.$\sqrt{s(s+2)}\dfrac{h}{2\pi }$
D.$\sqrt{(s+1)}\dfrac{h}{2\pi }$.

Answer 1

We know that 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 three directions have either the integer spin or the half integer spin. A spin quantum number is assigned to the elementary particles of the given kind that have the same spin angular momentum.

Complete answer:
An electron is believed to be a point particle and possessing no internal structure but still then the electron has spin. 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. Although the direction of the spin of the particle can be changed, it cannot be made to spin faster or slower.
Now, Orbital angular momentum is the component of angular momentum. It is the value of angular momentum of the electron revolving around the orbit and the fact that the electron is spinning around its own axis is neglected i.e. the spin is constant.
According to the conventional definition of the spin quantum number, $S=\dfrac{n}{2}$ where there can be any non-negative integer value. So, we get the allowed values of spin as $0,0.5,1,1.5$ etc. Thus, the spin is quantized and can take only discrete values. The elementary values are called Fermions whereas the elementary particles for which are called Bosons. The spin angular momentum S of any physical system is given as: $S=\hbar \sqrt{s(s+1)}=\dfrac{h}{2\pi }\sqrt{n(n+1)}=\sqrt{s(s+1)}\dfrac{h}{2\pi }$

Therefore, the correct answer is option A.

Note:
Remember that angular momentum of an electron was given by Bohr. This is the result of applying quantum theory to the orbit of the electron. The solution of the Schrodinger equation yields the angular momentum quantum number.