91影视

The Azimuthal quantum number specifies the shape and angular momentum of the orbital in the space.

Given that,

If $L_2$can take on only those values $m_{l}h$, where $m_l=0,卤1,鈥�鈥�卤l$, then its square is allowed only values $m{l}^2{h}^2$, and the average of $l^2$ should be the sum of its allowed values divided by the number of values, $2l+1$, because there really is no preferred direction in space, the averages of $L_x^{2}and{L}_y^2$ should be the same, and sum of all three should give the average of$L^2$ .

Where, L is the Orbital angular momentum, $L_x$ is the component of orbital angular momentum along x-axis, $L_y$ is the component of orbital angular momentum along y-axis, $L_z$ is the component of orbital angular momentum along z-axis, l is theAzimuthal quantum number,$m_l$ is theMagnetic quantum number.

role="math" localid="1659179529670" $L_z^2鈥�=鈥�\frac{1}{2l+1}\underset{ml=-l}{\overset{l}{鈭�}}m{l}^2{h}^2$

Where, h is Planck鈥檚 constant.

$$\begin{gathered} L_z^2鈥�=鈥�\frac{2}{2l+1}\underset{ml=-l}{\overset{l}{鈭�}}m{l}^2{h}^2 \\ =\frac{2}{2l+1}{h}^{2}l\left(l+1\right)\left(2l+1\right)/6 \\ =\frac{1}{3}{h}^{2}l\left(l+1\right) \end{gathered}$$

It is also given that,

$Avg.of{L}_x^2=Avg.of{L}_y^2=Avg.of{L}_z^2$ 鈥�.. (1)

Hence,

$$\begin{gathered} Avg.of{L}^2=Avg.of{L}_y^2=Avg.of{L}_z^2 \\ =3脳Avg.of{L}_x^2+Avg.of{L}_y^2+Avg.of{L}_z^2 \\ =3脳Avgof{L}_z^2 \\ =3脳\frac{1}{3}{h}^{2}l\left(l+1\right) \\ =h^{2}l\left(l+1\right) \end{gathered}$$