Answer to Question #118727 in Atomic and Nuclear Physics for Moel Tariburu

Question

Determine the total angular momentum quantum number if the minimum angle between the total angular momentum vector (in a single electron atom) and the z axis is 28.13°.

Answer 1

The minimum angle between the  total angular momentum vector (in a single electron atom) and the z axis is 

$cos\theta=\frac{l (\frac{h}{2\pi})}{\sqrt{l(l+1)}(\frac{h}{2\pi})}$

$cos28.13°=\frac{l (\frac{h}{2\pi})}{\sqrt{l(l+1)}(\frac{h}{2\pi})}$

$cos28.13°=\frac{l }{\sqrt{l(l+1)}}$

$0.88=\frac{l }{\sqrt{l(l+1)}}$

on solving the above equation, we get $l=3.43=3 (Approximately)$

So the total angular momentum is

$L=\sqrt{l(l+1)}(\frac{h}{2\pi})=\sqrt{3(3+1)}(\frac{h}{2\pi})=\sqrt{12}(\frac{h}{2\pi})$

$L=\sqrt{12}(\frac{h}{2\pi})=\sqrt{12}(\frac{6.626*10^{-34}}{2\pi})=3.65*10^{-34} J.s$