Question

The angular momentum of electron in ‘d’ orbital is equal to:
(A) $\dfrac{2\sqrt{3}h}{2\pi }$
(B) 0h
(C) $\dfrac{\sqrt{6}h}{2\pi }$
(D) $\dfrac{\sqrt{2}h}{2\pi }$

Answer 1

Hint: Recollect the concept of quantum numbers. Think about the azimuthal quantum number, l. Find out what angular momentum is. The formula for calculating angular momentum is, $L=\dfrac{h}{2\pi }\sqrt{l(l+1)}$. The question is to calculate angular momentum of d-orbital. Just substitute the value in the formula and find out the answer.

Complete answer:
- Azimuthal quantum number, l is the number which gives an idea in which subshell or orbital an electron is present. Azimuthal quantum number is a whole integer.
- For s-orbital, l=0.
- For p-orbital, l=1.
- For d-orbital, l=2.
- For f-orbital, l=3.
- Azimuthal quantum number is also known as orbital momentum quantum number or angular momentum quantum number.
- Angular momentum is given by the formula, $L=\dfrac{h}{2\pi }\sqrt{l(l+1)}$ where l is the azimuthal quantum number and h is the Planck’s constant.
- Now let’s take a look at the question. According to the question, we need to find angular momentum of d-orbital.
- For d-orbital, l=2. Therefore, angular momentum is given as,
\[L=\dfrac{h}{2\pi }\sqrt{2(2+1)}=\dfrac{\sqrt{6}h}{2\pi }\]
- Therefore, for d-orbital, angular momentum is equal to $\dfrac{\sqrt{6}h}{2\pi }$.

Therefore, the answer is option C.

Note:
Remember that azimuthal quantum number is known as orbital or angular momentum quantum number. The equation for calculating orbital angular momentum is $L=\dfrac{h}{2\pi }\sqrt{l(l+1)}$ where l is the azimuthal quantum number and h is the Planck’s constant.