Question

What is the orbital-angular momentum of an electron in ‘f’ orbital?
A. $\dfrac{{1.5\,h}}{\pi }$
B. $\dfrac{{\sqrt 6 \,h}}{\pi }$
C. $\dfrac{{\sqrt 3 \,h}}{\pi }$
D. $\dfrac{{\sqrt 3 \,h}}{{2\pi }}$

Answer 1

Hint: We know that angular momentum of an electron compromises of two parts: Orbital angular momentum (due to momentum of electron in the orbit with velocity v) and spin angular momentum (the electron spins in its own axis). The orbital angular momentum can be calculated using the formula $L = \sqrt {e(e + 1)} \,\dfrac{h}{{2\pi }}$.

Complete step by step answer:
The electron is in ‘f’ electron
The electron can have both orbital angular momentum as well as spin angular momentum. The electron is capable of revolving in the Bohr orbit and rotating on its own on its own axis, with velocity v.

The angular moment in the n Bohr Orbit is given as $\dfrac{{nh}}{{2\pi }}$, where n is the principal quantum number and h is the Planck's constant. The orbital angular momentum on other hand is given by L, where $L = \sqrt {e(e + 1)} \,\dfrac{h}{{2\pi }}$, l is the azimuthal quantum number or angular quantum number. Its value ranges from 1 to n-1. Also, its value gives shape of the orbital say s, p, d, f for 0, 1, 2, 3. Since the electron is in ‘f’ orbital it’s value will be 3. So, placing l=3 in the formula, we get
$L = \sqrt {3(3 + 1)} \,\dfrac{h}{{2\pi }}$
\[ \Rightarrow 2\sqrt 3 \,\dfrac{h}{\pi }\]
Therefore, $L = \,\sqrt 3 \dfrac{h}{\pi }$

Hence, option (C) is the correct answer.

Note:
The value of l for s, p , d , f orbitals is 0,1,2,3 respectively. Be careful while