91影视

The equation yields stationary-state wave functions for a particle in a three-dimensional cubical box;

${蠄}_{n_x,n_y{n}_z}(x,y,z)=C\sin 鈦�\left(\frac{n_X蟺x}{L}\right)\sin 鈦�\left(\frac{n_y蟺y}{L}\right)\sin 鈦�\left(\frac{n_z蟺z}{L}\right)$

As a result, the probability distribution function of a particle in a three-dimensional cubical box is as follows:

${\left|{\mathrm{唯}}_{n_x,n_y,n_z}(x,y,z)\right|}^2={\left(\frac{2}{L}\right)}^3{\sin }^2鈦�\left(\frac{n_X蟺x}{L}\right){\sin }^2鈦�\left(\frac{n_Y蟺y}{L}\right){\sin }^2鈦�\left(\frac{n_z蟺z}{L}\right)$

The probability distribution function for the state $\left(n_x,n_y,n_Z\right)=\left(2,2,1\right)$is;

${\left|{\mathrm{唯}}_{2,2,1}(x,y,z)\right|}^2={\left(\frac{2}{L}\right)}^3{\sin }^2鈦�\left(\frac{2蟺x}{L}\right){\sin }^2鈦�\left(\frac{2蟺y}{L}\right){\sin }^2鈦�\left(\frac{蟺z}{L}\right)$

when,

$$\begin{gathered} \sin 鈦�\left(\frac{2蟺x}{L}\right)=0鈬�\frac{2蟺x}{L}=0,蟺,2蟺 \\ x=0,\frac{L}{2},2L \\ \sin 鈦�\left(\frac{2蟺y}{L}\right)=0鈬�\frac{2蟺y}{L}=0,蟺,2蟺 \\ y=0,\frac{L}{2},L \\ \sin 鈦�\left(\frac{蟺z}{L}\right)=0鈬�\frac{蟺z}{L}=0,蟺,2蟺 \\ z=0,L \end{gathered}$$

The box's six walls are x= (0, L), y= (0, L), and z= (0, L). As a result, the other planes where ${\left|{蠄}_{2,2,1}\left(x,y,z\right)\right|}^2=0$exists are x=L/2 and y=L/2.

The probability distribution function for the state $\left(n_X,n_Y,n_Z\right)=\left(2,1,1\right)$is;

${\left|{蠄}_{2,1,1}(x,y,z)\right|}^2={\left(\frac{2}{L}\right)}^3{\sin }^2鈦�\left(\frac{2蟺x}{L}\right){\sin }^2鈦�\left(\frac{蟺y}{L}\right){\sin }^2鈦�\left(\frac{蟺z}{L}\right)$

when,

$$\begin{gathered} \sin 鈦�\left(\frac{2蟺x}{L}\right)=0鈬�\frac{2蟺x}{L}=0,蟺,2蟺 \\ x=0,\frac{L}{2},2L \\ \sin 鈦�\left(\frac{蟺y}{L}\right)=0鈬�\frac{2蟺y}{L}=0,蟺,2蟺 \\ y=0,L \\ \sin 鈦�\left(\frac{蟺z}{L}\right)=0鈬�\frac{蟺z}{L}=0,蟺,2蟺z=0,L \end{gathered}$$

As a result, there is only one other plane where ${\left|{\mathrm{唯}}_{2,1,1}(x,y,z)\right|}^2=0isx=\mathrm{L}/2$is x=L/2.

The probability distribution function for the state role="math" localid="1668240176081" $\left(n_X,n_Y,n_Z\right)=\left(1,1,1\right)$is;

${\left|{蠄}_{1,1,1}(x,y,z)\right|}^2={\left(\frac{2}{L}\right)}^3{\sin }^2鈦�\left(\frac{蟺x}{L}\right){\sin }^2鈦�\left(\frac{蟺y}{L}\right){\sin }^2鈦�\left(\frac{蟺z}{L}\right)$

when,

$$\begin{gathered} \sin 鈦�\left(\frac{蟺x}{L}\right)=0鈬�\frac{蟺x}{L}=0,蟺,2蟺 \\ x=0,L \\ \sin 鈦�\left(\frac{蟺y}{L}\right)=0鈬�\frac{蟺y}{L}=0,蟺,2蟺 \\ y=0,L \\ \sin 鈦�\left(\frac{蟺z}{L}\right)=0鈬�\frac{蟺z}{L}=0,蟺,2蟺 \\ z=0,L \end{gathered}$$

As a result, there are no other planes where ${\left|{蠄}_{1,1,1}(x,y,z)\right|}^2=0$can exist.

Hence,

There are two nodal planes in the $\left(n_X,n_Y,n_Z\right)=\left(2,2,1\right)$state.

There is only one nodal plane in the $\left(n_X,n_Y,n_Z\right)=\left(2,1,1\right)$states.

There is no nodal plane in$\left(n_X,n_Y,n_Z\right)=\left(2,2,1\right)$ states.

As a result, as energy increases, so does the number of nodal planes.