Question

The angular momentum of an electron in 4s orbital, 3p orbital and 4th orbit respectively are:
A. $0,\dfrac{h}{{\sqrt 2 \pi }},\dfrac{{2h}}{\pi }$
B.$\dfrac{1}{{\sqrt 2 }},\dfrac{h}{\pi },\dfrac{{2h}}{\pi },0$
C. $0,\dfrac{{\sqrt 2 h}}{\pi },\dfrac{{4h}}{\pi }$
D. $\dfrac{{\sqrt 2 h}}{\pi },\dfrac{{4h}}{\pi },0$

Answer 1

Hint: -Using the formula for orbital angular momentum L = $\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$ we can easily find the angular momentum of the electrons in certain orbitals and for orbits using L =$\dfrac{{nh}}{{2\pi }}$

Complete step by step answer:
Now from the question
Angular momentum of electron in 4s orbital =$\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
For s orbital $l$ =0
Therefore Angular momentum =$\sqrt {0(0 + 1)} \dfrac{h}{{2\pi }}$= 0
Angular momentum of electron in 3p orbital =$\sqrt {l(l + 1)} \dfrac{h}{{2\pi }}$
For p orbital $l$ = 1
Therefore Angular momentum = $\sqrt {1(1 + 1)} \dfrac{h}{{2\pi }}$=$\dfrac{h}{{\sqrt 2 \pi }}$

Angular momentum of electron in 4th orbit =$\dfrac{{nh}}{{2\pi }}$
Here n = 4
Therefore Angular momentum =$\dfrac{{4h}}{{2\pi }}$ = $\dfrac{{2h}}{\pi }$
Therefore the angular momentums of electron in 4s orbital, 3p orbital and 4th orbit are 0 , $\dfrac{h}{{\sqrt 2 \pi }}$, $\dfrac{{2h}}{\pi }$
So, the correct answer is “Option A”.

Note: In the process of solving the Schrodinger equation for the atom , it's found that the orbital momentum is quantized consistent with the connection .
${L^2} = l(l + 1){(\dfrac{h}{{2\pi }})^2}$
It is a characteristic of angular momenta in quantum physics that the magnitude of the momentum in terms of the orbital quantum number is of the shape.
 L =$\dfrac{{nh}}{{2\pi }}$