91Ó°ÊÓ

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

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

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

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

The probability of detection at $\left(x,y\right)$is given by,

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

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

$\mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{Ï€}}{\mathrm{L}}\mathrm{x}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{Ï€}}{\mathrm{L}}\mathrm{y}\right)\mathrm{dxdy}$

It is given that the probe is placed at (x,y) = ( 0.200L, 0.800L ). Therefore,

$$\begin{gathered} \mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{x}}\mathrm{Ï€}}{\mathrm{L}}\left(0.200L\right)\right){\sin }^2\left(\frac{{\mathrm{n}}_{\mathrm{y}}\mathrm{Ï€}}{\mathrm{L}}\left(0.800L\right)\right)\mathrm{dxdy} \\ =\left(\frac{4}{{\mathrm{L}}^2}\right){\sin }^2\left(0.200{\mathrm{n}}_{\mathrm{x}}\mathrm{Ï€}\right){\sin }^2\left(0.800{\mathrm{n}}_{\mathrm{y}}\mathrm{Ï€}\right)\mathrm{dxdy} \end{gathered}$$

Substitute $150\mathrm{pm}\mathrm{for}\mathrm{L},5.00\mathrm{pm}\mathrm{for}\mathrm{dx},5.00\mathrm{pm}\mathrm{for}\mathrm{dy},1\mathrm{b}\mathrm{for}{\mathrm{n}}_{\mathrm{x}},\mathrm{and}3\mathrm{for}{\mathrm{n}}_{\mathrm{y}}$in the above probability formula.

$$\begin{gathered} \mathrm{p}\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{4}{{\left(150\mathrm{pm}\right)}^2}\right){\sin }^2\left(0.200×1×\mathrm{π}\right){\sin }^2\left(0.800×3×\mathrm{π}\right){\left(5.00\mathrm{pm}\right)}^2 \\ =\left(1.77×{10}^{-4}{\mathrm{pm}}^{-2}\right){\left(0.5878\right)}^2{\left(0.951\right)}^2\left(25.00{\mathrm{pm}}^2\right) \\ =\left(1.77×{10}^{-4}{\mathrm{pm}}^{-2}\right){\left(0.0105\right)}^2{\left(0.131\right)}^2\left(25.00{\mathrm{pm}}^2\right) \\ =1.4×{10}^{-3} \end{gathered}$$

Therefore, the required probability of detection is $1.4×{10}^{-3}$.