Question

Angular momentum for P-shell electron is:
A. $\dfrac{\text{3h}}{\text{ }\!\!\Pi\!\!\text{ }}$
B. Zero
C. $\dfrac{\sqrt{\text{2}}\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$
D. None

Answer 1

Hint: Bohr gave the postulates to define the angular momentum of the electrons. Angular momentum is represented by L and the formula used to calculate angular momentum is $\text{L=}\sqrt{\text{l(l+1)}}\text{.}\dfrac{\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$ , where l is azimuthal quantum number.

Complete step by step solution:
As we know that whenever any quantity rotates it poses some momentum, this momentum is called angular momentum. We know that electrons revolve in their shell, so they also possess some angular momentum. Angular momentum can be calculated by using the formula $\text{L=}\sqrt{\text{l(l+1)}}\text{.}\dfrac{\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$ where l is azimuthal quantum number. The value of l ranges from 0 to n-1 and n represents the shell.
For s shell value of n is zero.
For p shell value of n is 1.
For d the shell value of n is 2.
For f shell value of n is 3.
In question it is given that electron is present in p shell therefore, its angular momentum will be: $\text{L=}\sqrt{\text{l(l+1)}}\text{.}\dfrac{\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$, put the value of l = 1 we get $\dfrac{\sqrt{\text{2}}\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$.

Therefore, option C is correct.

Additional Information:
Angular momentum can also be calculated as mvr, where m is the mass of the electron, v is the velocity of the electron and r is the radius of the electron in which it is revolving.

Note: The angular momentum of s orbital is zero. It is zero because the value of l for s shell is zero and if we put l = 0 in the formula $\text{L=}\sqrt{\text{l(l+1)}}\text{.}\dfrac{\text{h}}{\text{2 }\!\!\Pi\!\!\text{ }}$ we get angular momentum as zero. Zero angular momentum means s orbitals are spherically symmetrical.