For a $p$ electron, the orbital angular momentum is:
A. $\dfrac{\sqrt{6}h}{2\pi }$
B. $\dfrac{\sqrt{2}h}{2\pi }$
C. $\dfrac{h}{2\pi }$
D. $\dfrac{2h}{2\pi }$

Hint: The orbital angular momentum of an electron is an integral multiple of $\dfrac{h}{2\pi }$ as explained by Bohr’s atomic theory model. The integral number depends on the azimuthal quantum number $\left( l \right)$ of the electron. The azimuthal quantum numbers of s,p,d and f electrons are 0,1,2 and 3 respectively.

Formula used:
The orbital angular momentum of an electron with azimuthal quantum number $l$ is given by
$\sqrt{l\left( l+1 \right)}\dfrac{h}{2\pi }$
Where h is the planck's constant equal to $6.626\times {{10}^{-34}}J.s$

Complete Step-by-Step solution:
As the electron revolves in an orbital around the nucleus according to the Bohr atomic theory, it possesses an orbital angular momentum. This orbital angular momentum depends on the azimuthal quantum number $\left( l \right)$ of the electron. The azimuthal quantum numbers of s,p,d and f electrons are 0,1,2 and 3 respectively.
The orbital angular momentum of an electron with azimuthal quantum number $l$ is given by
$\sqrt{l\left( l+1 \right)}\dfrac{h}{2\pi }$ ---(1)
Where h is the planck's constant equal to $6.626\times {{10}^{-34}}J.s$
The azimuthal quantum number $\left( l \right)$ of a $p$ electron is $l=1$ . Hence, to find the orbital angular momentum of a $p$ electron, we have to put this value in equation (1),
Hence, doing so, we get,
Orbital angular momentum of $\text{p} \text{ electron = }\sqrt{1\left( 1+1 \right)}\dfrac{h}{2\pi }=\sqrt{1\left( 2 \right)}\dfrac{h}{2\pi }=\dfrac{\sqrt{2}h}{2\pi }$.
Hence, the required value of the orbital angular momentum is $\dfrac{\sqrt{2}h}{2\pi }$.
Hence, the correct option is B) $\dfrac{\sqrt{2}h}{2\pi }$.

Note: Due to the fact that an electron carries charge and revolves in a circular path around the nucleus according to the Bohr atomic theory, it also acts like a current carrying circular coil and hence also produces its own magnetic field. This can be measured in Bohr magneton.
It is almost obvious that the formula (1) is correct since d and f electrons which are situated further away from the nucleus must have a larger orbital angular momentum from simple mechanics. This is also true by the formula since d and f electrons have azimuthal quantum number values of 2 and 3 respectively which are more than that of s and p electrons (0 and 1 respectively).