Chapter 8, Problem 8.1CYU

Check Your Understanding Can the magnitude of L_zever be equal to L?

To determine

Whether the magnitude of $L_z$ can be equal to $L$.

Answer to Problem 8.1CYU

The angular moment of $z$ component $\left(L_z\right)$ is equal to its angular moment $\left(L\right)$ when the angle between $L_z-L$ is $0^0$. $$

Explanation of Solution

Introduction:

The angular moment $\left(L\right)$ possess by the electron is a product of distance of the electron from the center of rotation and the momentum of the electron. The product of linear velocity and mass of electron gives the quantity of momentum.

  $L_z$ is the $z$ component orbital angular momentum and expressed as $L_z=m\hslash$.

Here, m indicates the angular momentum projection quantum number and $\hslash$ indicates the reduced Plank's constant.

Explantion:

The relation for an angular moment $\left(L\right)$ and orbital angular momentum for $z$ the component $\left(L_z\right)$ is

  $L_z=L\cos \theta$ ...... (1)

Here, $\theta$ indicates an angle between $z$ - axis and angular momentum vector.

In order for the value of $L_z$ and $L$ to be equal the value of $\theta$ must be $0$, as the value of $\text{cos}\left(0\right)=1$.

Substitute $0$ for $\theta$ in equation (1),

  $\begin{array}{l}{L}_z=L\cos 0\\ L_z=L\end{array}$

Hence, for the magnitude of $L_z$ and $L$ to be equal the angle between these two components must be $0$.

Figure A.

The above figure shows the movement of the electron and its angular moment along the z -axis. The notation r, p, and L indicate the distance of the electron from the center axis the momentum of the electron and angular moment of the electron, which overlaps the angular moment of z - component.

Conclusion:

The angular moment of $z$ component $\left(L_z\right)$ is equal to its angular moment $\left(L\right)$ when the angle between $L_z-L$ is $0^0$. $$