Question

The orbital angular momentum of a 4p electron will be
(A) $4.\dfrac{h}{2\pi }$
(B) $\sqrt{2}.\dfrac{h}{2\pi }$
(C) $\sqrt{6}.\dfrac{h}{4\pi }$
(D) $\sqrt{2}.\dfrac{h}{4\pi }$

Answer 1

Orbital angular momentum is the speed of an electron revolving around the nucleus for one complete path or rotation. It depends on the value of azimuthal quantum number as it describes the shape of the orbital.

Complete Solution :
Let us first learn about the concepts related to orbital angular momentum and then move towards the given illustration:
Angular momentum-
It is rotational equivalent of the linear momentum i.e. it is a measure of the momentum of a body in rotational motion.
Orbital angular momentum-
It is a component of angular momentum. Orbital angular momentum is the value of angular momentum of the electron revolving around the orbit where the spin of an electron around its axis is neglected.
It is given as:
$OAM=\dfrac{\sqrt{l\left( l+1 \right)}\times h}{2\pi }$
where,
l = azimuthal quantum number
h = planck's constant
Now,
Illustration-
The orbital angular momentum of a 4p electron is given as;
The azimuthal quantum number for 4p orbital is 1 i.e. l = 1.
We have:
$OAM=\dfrac{\sqrt{l\left( l+1 \right)}\times h}{2\pi }$
putting the values described we get:

& OAM=\dfrac{\sqrt{1\left( 1+1 \right)}\times h}{2\pi } \\\ & OAM=\dfrac{\sqrt{2}\times h}{2\pi } \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** Here, if the option as $\dfrac{h}{\sqrt{2}\times \pi }$ also, will be correct. Thus, do not get confused if any option is the same as described above as; $OAM=\dfrac{h\sqrt{2}}{2\pi }=\dfrac{h}{\sqrt{2}\times \pi }$ .