Question

Spin angular momentum is closely analogous to the treatment of orbital angular momentum.
Spin angular momentum = \[\sqrt {s(s + 1)h} \]
Orbital angular momentum = \[\sqrt {l(l + 1)h} \]
Total spin of an atom or ion is a multiple of \[\dfrac{1}{2}\]. Spin multiplicity is a factor to confirm the electronic configuration of an atom or ion. Spin multiplicity = \[{\text{(2}}\sum {\text{s}} {\text{ + 1)}}\]\[\]
The orbital angular momentum for a 2-p electron is:
(A) $\sqrt {3h} $
(B) $\sqrt {6h} $
(C) $Zero$
(D) $\sqrt 2 \dfrac{h}{{2\pi }}$

Answer 1

Hint: To answer this question, we should know that angular momentum is rotational equivalent to linear momentum. Linear momentum is the product of a system's mass multiplied by its velocity whereas, angular momentum (L) is the distance of the object from a rotation axis multiplied by the linear momentum

Complete step by step solution:
-First of all, we will discuss spin quantum numbers.
We should know that the spin quantum number (\[{{\text{m}}_{\text{s}}}\]) basically describes the angular momentum of an electron. An electron spinning around an axis will have both angular momentum and orbital angular momentum. We know that angular momentum is a vector quantity and also that the Spin Quantum Number (s) has a magnitude of (1/2) and direction (+ or -).
-We will now talk about orbital angular momentum.
Angular momentum comprises orbital and spin angular momentum and so orbital angular momentum is said to be a component of angular momentum. It is basically the value of the angular momentum of the electron revolving around in its own orbit and we can say that the spinning of the electron around its own axis is neglected.
-Now, coming back to our question we will calculate the orbital angular momentum for a 2-p electron.
We know that orbital angular momentum= $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$ (1)
We also know that for a p-orbital the value of l = 1.
So, to find out the value of orbital angular momentum we will put the value of l = 1 for a p-orbital in the equation (1):
Orbital angular momentum= $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
$ = \sqrt {1(1 + 1)} \dfrac{h}{{2\pi }}$
$ = \sqrt 2 \dfrac{h}{{2\pi }}$
Hence we obtain the value of orbital angular momentum for p-orbital to be \[ = \sqrt 2 \dfrac{h}{{2\pi }}\]

So, the correct option will be: (D) $\sqrt 2 \dfrac{h}{{2\pi }}$.

Note: In an atom there are a total of four quantum numbers: the principal quantum number (n) which represents the shell number or the energy level, the orbital angular momentum quantum number (l) which tells us about the orbital shape, the magnetic quantum number (${m_l}$) which tells us about the orientation of the orbital and the electron spin quantum number (${m_s}$) which represents the electron spin direction.