Question

Figure 39-30 shows a two-dimensional, infinite-potential well lying in an xy plane that contains an electron. We probe for the electron along a line that bisects ${\mathbf{L}}_{\mathbf{x}}$and find three points at which the detection probability is maximum. Those points are separated by 2.00 nm . Then we probe along a line that bisects ${\mathbf{L}}_{\mathbf{y}}$and find five points at which the detection probability is maximum. Those points are separated by 3.00 nm . What is the energy of the electron?

Short answer

The energy of the electron is 0.136 eV.

Step-by-step solution

The energy of two-dimensional electron traps:

The quantized energies for an electron trapped in a two-dimensional infinite potential well that forms a rectangular corral are,

$\mathbf{E}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$ ….. (1)

Here, ${\mathbf{n}}_{\mathbf{x}}$is quantum number for which the electron’s matter wave fits in well width ${\mathbf{L}}_{\mathbf{x}}$, ${\mathbf{n}}_{\mathbf{y}}$is quantum number for which the electron’s matter wave fits in well width ${\mathbf{L}}_{\mathbf{y}}$, h is plank constant, and is mass of the electron.

Find the energy of the electron:

Every probability maximum represents a quantum number, in this case in x- direction, $n_x=3$and in y-direction $n_y=5$.

Substitute localid="1661774839620" $6.626\times {10}^{-34}\hspace{0.33em}\mathrm{J}\cdot \mathrm{s}\mathrm{for}\mathrm{h},9.109\times {10}^{-31}\mathrm{kg}\mathrm{for}\mathrm{m},3\mathrm{for}{\mathrm{n}}_{\mathrm{x}},5\mathrm{for}{\mathrm{n}}_{\mathrm{y}},2\times {10}^{-9}\hspace{0.33em}\mathrm{m}\mathrm{for}{\mathrm{L}}_{\mathrm{x}}.\mathrm{and}3\times {10}^{-9}\mathrm{m}\mathrm{for}{\mathrm{L}}_{\mathrm{y}}$in equation (1).

$$\begin{gathered} \mathrm{E}=\frac{{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{\left(8\right)\left(9.109\times {10}^{-31}\mathrm{kg}\right)}\left(\frac{3^3}{{\left(3.2\times {10}^{-9}\mathrm{m}\right)}^2}+\frac{5^2}{{\left(5.3\times {10}^{-9}\mathrm{m}\right)}^2}\right) \\ =2.18\times {10}^{-20}\mathrm{J} \\ =\left(2.18\times {10}^{-20}\mathrm{J}\right)\left(\frac{6.242\times {10}^{18}\mathrm{eV}}{1\mathrm{J}}\right) \\ =0.136\mathrm{eV} \end{gathered}$$

Hence, the energy of the electron is 0.136 eV .