91影视

In a three-dimensional cubical box, the energy level of a particle is written as:

${\mathit{E}}_{{\mathbf{n}}_{\mathbf{X}}\mathbf{,}{\mathbf{n}}_{\mathbf{Y}}\mathbf{,}{\mathbf{n}}_{\mathbf{z}}}\mathbf{=}\frac{\left(n_X^2+n_Y^2+n_Z^2\right){\mathbf{蟺}}^{\mathbf{2}}{\mathbf{魔}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}$鈥︹赌︹赌︹赌︹赌�(1)

Where,${\mathit{n}}_{\mathbf{X}}\mathbf{,}{\mathit{n}}_{\mathbf{Y}}$ and${\mathit{n}}_{\mathbf{z}}$ are the quantum numbers.

In a three-dimensional cubical box, the stationary-state wave function for a particle is represented as:

${\left|{唯}_{n_x,n_y,n_z}(x,y,z)\right|}^{\mathbf{2}}\mathbf{=}{\left(\frac{2}{L}\right)}^{\mathbf{3}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{鈦�}\frac{{\mathbf{n}}_{\mathbf{x}}\mathbf{蟺}\mathbf{x}}{\mathbf{L}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{鈦�}\frac{{\mathbf{n}}_{\mathbf{纬}}\mathbf{蟺}\mathbf{y}}{\mathbf{L}}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{鈦�}\frac{{\mathbf{n}}_{\mathbf{z}}\mathbf{蟺}\mathbf{z}}{\mathbf{L}}$

Given that, the cubical volume is defined by$0鈮�x鈮�L/4,0鈮�y鈮�L/4$ and$0鈮�z鈮�L/4$

(a)

The volume of a three-dimensional cubical box with length L/4 is:

$$\begin{gathered} V_{L/4}={\left(\frac{L}{2}\right)}^3 \\ =\frac{L^3}{64} \end{gathered}$$

As the volume of the three-dimensional cubical box is $\mathrm{V}={\mathrm{L}}^3$.

A fraction of the total volume of the box is:

$$\begin{gathered} fraction=\frac{V_{L/4}}{L^3} \\ =\frac{L^3/64}{L^3} \\ =0.0156 \end{gathered}$$

The probability of particles in the cubical volume:

$$\begin{gathered} \mathrm{p}={鈭�}_0^{{\mathrm{L}}^{3/64}}鈥�{\left|{蠄}_{{\mathrm{n}}_x,{\mathrm{n}}_Y,{\mathrm{n}}_{\mathrm{z}}}(\mathrm{x},\mathrm{y},\mathrm{z})\right|}^2\mathrm{dV} \\ ={鈭�}_0^{L^3/64}鈥�\left[{\left(\frac{2}{\mathrm{L}}\right)}^3{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{x}}蟺\mathrm{x}}{\mathrm{L}}{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{Y}}蟺\mathrm{y}}{\mathrm{L}}{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{z}}蟺\mathrm{z}}{\mathrm{L}}\right]\mathrm{dV} \\ ={\left(\frac{2}{\mathrm{L}}\right)}^3\left[{鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{X}}蟺\mathrm{x}}{\mathrm{L}}\mathrm{dx}\right]\left[{鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{Y}}蟺\mathrm{y}}{\mathrm{L}}\mathrm{dy}\right]\left[{鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{z}}蟺\mathrm{Z}}{\mathrm{L}}\mathrm{dz}\right] \\ ={\left(\frac{2}{\mathrm{L}}\right)}^3{\mathrm{I}}_{\mathrm{x}}{\mathrm{I}}_{\mathrm{y}}{\mathrm{I}}_{\mathrm{z}} \end{gathered}$$

Hence, the fraction of the total volume is 0.0156.

(b)

For the ground state$(n_X,n_Y,n_Z)=(1,1,1)$the x,y andintegrals are:

$$\begin{gathered} {\mathrm{I}}_{\mathrm{x}}={鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{n_x蟺x}{L}dx \\ =\frac{L}{8}鈭�\frac{L}{4n_x蟺}\sin 鈦�\frac{n_x蟺}{2} \\ =\frac{L}{8}鈭�\frac{L}{4蟺} \\ =\left(\frac{L}{2}\right)\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right) \end{gathered}$$

$$\begin{gathered} I_Y={鈭�}_0^{L/4}鈥�{\sin }^2鈦�\frac{n_Y蟺y}{L}dy \\ =\frac{L}{8}鈭�\frac{L}{4n_Y蟺}\sin 鈦�\frac{n_Y蟺}{2} \\ =\frac{L}{8}鈭�\frac{L}{4蟺} \\ =\left(\frac{L}{2}\right)\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_{\mathrm{z}}={鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{X}}蟺\mathrm{z}}{\mathrm{L}}\mathrm{dz} \\ =\frac{\mathrm{L}}{8}鈭�\frac{\mathrm{L}}{4{\mathrm{n}}_{\mathrm{z}}蟺}\sin 鈦�\frac{{\mathrm{n}}_{\mathrm{z}}蟺}{2} \\ =\frac{\mathrm{L}}{8}鈭�\frac{\mathrm{L}}{4蟺} \\ =\left(\frac{\mathrm{L}}{2}\right)\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right) \end{gathered}$$

The probability of the particle in the ground state is:

$$\begin{gathered} p_{1,1,1}={\left(\frac{2}{L}\right)}^3{\left(\frac{L}{2}\right)}^3{\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right)}^3 \\ ={\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right)}^3 \\ =7.50脳{10}^{鈭�4} \end{gathered}$$

Hence, the probability of the particle in the ground state will be the cubical volume defined in part (a) is $7.50脳{10}^{-4}$.

(c)

For the state$(n_X,n_Y,n_Z)=(2,1,1)$ the x,y andintegrals are:

$$\begin{gathered} {\mathrm{I}}_{\mathrm{x}}={鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{x}}蟺x}{\mathrm{L}}\mathrm{dx} \\ =\frac{L}{8}鈭�\frac{\mathrm{L}}{4{\mathrm{n}}_x蟺}\sin 鈦�\frac{{\mathrm{n}}_x蟺}{2} \\ =\frac{L}{8}鈭�\frac{L}{8蟺}\left(0\right) \\ =\left(\frac{L}{2}\right){\left(\frac{1}{2}\right)}^2 \end{gathered}$$

$$\begin{gathered} I_Y={鈭�}_0^{L/4}鈥�{\sin }^2鈦�\frac{n_Y蟺y}{L}dy \\ =\frac{L}{8}鈭�\frac{L}{4n_Y蟺}\sin 鈦�\frac{n_Y蟺}{2} \\ =\frac{L}{8}鈭�\frac{L}{4蟺} \\ =\left(\frac{L}{2}\right)\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{I}}_z={鈭�}_0^{\mathrm{L}/4}鈥�{\sin }^2鈦�\frac{{\mathrm{n}}_{\mathrm{x}}蟺\mathrm{z}}{\mathrm{L}}\mathrm{dz} \\ =\frac{\mathrm{L}}{8}鈭�\frac{\mathrm{L}}{4{\mathrm{n}}_z蟺}\sin 鈦�\frac{{\mathrm{n}}_z蟺}{2} \\ =\frac{\mathrm{L}}{8}鈭�\frac{\mathrm{L}}{4蟺} \\ =\left(\frac{\mathrm{L}}{2}\right)\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right) \end{gathered}$$

The probability of the particle in the ground state is:

$$\begin{gathered} {\mathrm{p}}_{2,1,1}={\left(\frac{2}{\mathrm{L}}\right)}^3{\left(\frac{\mathrm{L}}{2}\right)}^3{\left(\frac{1}{4}鈭�\frac{1}{2蟺}\right)}^2{\left(\frac{1}{2}\right)}^2 \\ ={\left(\frac{1}{8}鈭�\frac{1}{2蟺}\right)}^2 \\ =0.0021 \end{gathered}$$

Hence, the probability of the when the particle is in the state$n_X=2,n_Y=1,n_Z=1$ is 0.021.