What is the greatest magnitude of the orbital angular momentum L that you can find in a state with n = 6 ?
A) 5.48 $\frac { h } { 2 \pi }$
B) 5.92 $\frac { h } { 2 \pi }$
C) 6 $\frac { h } { 2 \pi }$
D) 6.48 $\frac { h } { 2 \pi }$

Question

Quizplus · Accepted Answer

The greatest magnitude of the orbital angular momentum L for a given principal quantum number n is given by $ L = \sqrt{l(l+1)} \frac{h}{2\pi} $, where l is the azimuthal quantum number and can take values from 0 to n-1. For n = 6, the maximum l is 5. Thus, $ L = \sqrt{5(5+1)} \frac{h}{2\pi} = \sqrt{30} \frac{h}{2\pi} \approx 5.48 \frac{h}{2\pi} $.

What Is the Greatest Magnitude of the Orbital Angular Momentum h2π\frac { h } { 2 \pi }2πh​

What Is the Greatest Magnitude of the Orbital Angular Momentum $\frac { h } { 2 \pi }$