Question

An electron is in the ground state in a two-dimensional, square, infinite potential well with edge lengths L. We will probe for it in a square of area $\mathbf{400}\mathbf{}{\mathbf{pm}}^{\mathbf{2}}$that is centered at $\mathbf{x}\mathbf{=}\mathbf{L}\mathbf{/}\mathbf{8}\mathbf{}\mathbf{and}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{L}\mathbf{/}\mathbf{8}$. The probability of detection turns out to be $\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}$. What is edge length L?

Short answer

The edge length is 27.6 nm.

Step-by-step solution

Describe the wave functions for an electron in a two-dimensional well:

The wave functions for an electron in a two-dimensional well are given by,

$$\begin{gathered} {\mathbf{\psi }}_{{\mathbf{n}}_{\mathbf{x}}\mathbf{,}{\mathbf{n}}_{\mathbf{y}}}\mathbf{}\left(x,y\right)\mathbf{=}{\mathbf{\psi }}_{{\mathbf{n}}_{\mathbf{x}}}\mathbf{}\left(x\right){\mathbf{\psi }}_{{\mathbf{n}}_{\mathbf{y}}}\mathbf{}\left(y\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{\frac{\mathbf{2}}{{\mathbf{L}}_{\mathbf{x}}}}\mathbf{sin}\left(\frac{n_x\pi }{L_x}x\right)\sqrt{\frac{\mathbf{2}}{{\mathbf{L}}_{\mathbf{y}}}}\mathbf{sin}\left(\frac{n_y\pi }{L_y}y\right) \end{gathered}$$

Find the edge length L:

In this case, the well is square, so ${\mathrm{L}}_{\mathrm{x}}={\mathrm{L}}_{\mathrm{y}}=\mathrm{L}$. Therefore,

$$\begin{gathered} {\mathrm{\psi }}_{{\mathrm{n}}_{\mathrm{x}},{\mathrm{n}}_{\mathrm{y}}}\left(\mathrm{x},\mathrm{y}\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{\pi }}{\mathrm{L}}\mathrm{x}\right)\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{\pi }}{\mathrm{L}}\mathrm{y}\right) \\ =\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{\pi }}{\mathrm{L}}\mathrm{x}\right)\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{\pi }}{\mathrm{L}}\mathrm{y}\right) \end{gathered}$$

The probability of detection at ( x , y ) is given by,

$\mathrm{p}\left(\mathrm{x},\mathrm{y}\right)={\left|{\mathrm{\psi }}_{{\mathrm{n}}_{\mathrm{x}},{\mathrm{n}}_{\mathrm{y}}}\left(\mathrm{x},\mathrm{y}\right)\right|}^2\mathrm{dxdy}$

Where, dx and are the width and height of the detection area, which are the width and height of the probe.

role="math" localid="1661774763899" $\mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{\pi }}{\mathrm{L}}\mathrm{x}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{\pi }}{\mathrm{L}}\mathrm{y}\right)\mathrm{dxdy}$

It is given that the probe is placed at $\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{L}}{8},\frac{\mathrm{L}}{8}\right)$. Therefore,

$$\begin{gathered} \mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{\pi }}{\mathrm{L}}\left(\frac{L}{8}\right)\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{\pi }}{\mathrm{L}}\left(\frac{L}{8}\right)\right)\mathrm{dxdy} \\ =\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{\pi }}{\mathrm{L}}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{\pi }}{\mathrm{L}}\right)\mathrm{dxdy} \end{gathered}$$

At the ground state,$\left({\mathrm{n}}_{\mathrm{x}},{\mathrm{n}}_{\mathrm{y}}\right)=\left(1,1\right)$So,

$$\begin{gathered} \mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{\mathrm{\pi }}{8}\right)\mathrm{dxdy} \\ =\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{\mathrm{\pi }}{8}\right)\mathrm{dxdy} \end{gathered}$$

Simplify for L as follow.

$$\begin{gathered} {\mathrm{L}}^2=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{\mathrm{\pi }}{8}\right)\mathrm{dxdy} \\ \mathrm{L}=2\sqrt{\frac{\mathrm{dxdy}}{\mathrm{p}}}{\sin }^2\left(\frac{\mathrm{\pi }}{8}\right) \end{gathered}$$

Substitute, $400{\mathrm{pm}}^2\mathrm{for}\mathrm{dxdy},4.5\times {10}^{-8}\mathrm{for}\mathrm{p}$in the above equation.

$$\begin{gathered} \mathrm{L}=2\sqrt{\frac{400}{4.5\times {10}^{-8}}}{\sin }^2\left(\frac{\mathrm{\pi }}{8}\right) \\ =2\sqrt{88.89\times {10}^8}\times \sin \left(\frac{\mathrm{\pi }}{8}\right)\sin \left(\frac{\mathrm{\pi }}{8}\right) \\ =2\times 9.428\times {10}^4\times 0.383\times 0.383 \\ \mathrm{L}=2.76\times {10}^4\mathrm{pm} \\ =27.6\mathrm{nm} \end{gathered}$$

Hence, the edge length is 27.6 nm.