The orbitals angular momentum for an electron revolving in an orbit is given by $\sqrt {l(l + l)} .\dfrac{h}{{2\pi }}.$ This momentum for an electron will be given by (where\[n = 2\]):

(A) zero
(B) $\dfrac{h}{{\sqrt 2 \pi }}$
(C) $\sqrt 2 .\dfrac{h}{\pi }$
(D) $ - \dfrac{1}{2}.\dfrac{h}{{2\pi }}$

Hint: Orbital angular momentum of an electron of an atom is arising due to motion around nuclear. It does not occur due to spin of electrons. Since I know electrons spin in either clockwise or anticlockwise direction while moving around the nucleus.

Step by step answer: As we know that orbital angular momentum is $ = \sqrt {l(l + l)} .\dfrac{h}{{2\pi }}$
Where $l = $azimuthal quantum number.
Azimuthal quantum number for,
S orbital = 0
P orbital = 1
D orbital = 2
F orbital =3
For a common revolving electron orbital angular momentum in given orbital $[n = 1,2,3,......]$
$mvr = n\dfrac{h}{{2\pi }}$
Where, \[m = \]mass of electron
\[v = \]velocity of electron around nucleus
\[r = \]radius of orbit
\[n = \]principle quantum number
\[h = \]planck's constant
$\therefore $angular momentum in h-orbital$ = \dfrac{{nh}}{{2\pi }}$
Since, $n = 2$is given
$ = 2.\dfrac{h}{{2\pi }}$
$ = \dfrac{h}{\pi }$
Therefore, from the above explanation the correct option is (C)$\sqrt 2 .\dfrac{h}{\pi }$
Since angular momentum depends an azimuthal quantum number $l.$
$\therefore $Angular momentum for 3d and 4d orbital are the same.
Because here principle quantum number changes but azimuthal quantum for both is the same.
Azimuthal quantum number for d-orbital$ = 2.$
$\therefore $Angular momentum$ = \sqrt {l(l + 1)} .\dfrac{h}{{2\pi }}$
$ = \sqrt {2(2 + 1)} .\dfrac{h}{{2\pi }}$
$ = \sqrt 6 .\dfrac{h}{{2\pi }}$
Same for 3d and 4d orbitals.

Hence option is C.

Note: There is an orbital angular momentum associated with an electron in a subshell, for each subshell we should compute it. We get this term because of the movement of electrons along circular paths.