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 quantum state is given by $L = \sqrt{l(l+1)} \frac{h}{2\pi}$, where $l$ is the azimuthal quantum number, which can take values from $0$ to $n-1$. For $n = 6$, the maximum value of $l$ is $5$. Substituting $l = 5$ into the formula gives $L = \sqrt{5(5+1)} \frac{h}{2\pi} = \sqrt{30} \frac{h}{2\pi} \approx 5.48 \frac{h}{2\pi}$.

What Is the Greatest Magnitude of the Orbital Angular Momentum n=6?n = 6 ?n=6?

What Is the Greatest Magnitude of the Orbital Angular Momentum $n = 6 ?$