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Expert-verified Found in: Page 1216 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # 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? The energy of the electron is 0.136 eV.

See the step by step solution

## Step 1: 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.

## Step 2: 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" $6.626×{10}^{-34} \mathrm{J}\cdot \mathrm{s}\mathrm{for}\mathrm{h},9.109×{10}^{-31}\mathrm{kg}\mathrm{for}\mathrm{m},3\mathrm{for}{\mathrm{n}}_{\mathrm{x}},5\mathrm{for}{\mathrm{n}}_{\mathrm{y}},2×{10}^{-9} \mathrm{m}\mathrm{for}{\mathrm{L}}_{\mathrm{x}}.\mathrm{and}3×{10}^{-9}\mathrm{m}\mathrm{for}{\mathrm{L}}_{\mathrm{y}}$in equation (1).

$\mathrm{E}=\frac{{\left(6.626×{10}^{-34}\mathrm{J}.\mathrm{s}\right)}^{2}}{\left(8\right)\left(9.109×{10}^{-31}\mathrm{kg}\right)}\left(\frac{{3}^{3}}{{\left(3.2×{10}^{-9}\mathrm{m}\right)}^{2}}+\frac{{5}^{2}}{{\left(5.3×{10}^{-9}\mathrm{m}\right)}^{2}}\right)\phantom{\rule{0ex}{0ex}}=2.18×{10}^{-20}\mathrm{J}\phantom{\rule{0ex}{0ex}}=\left(2.18×{10}^{-20}\mathrm{J}\right)\left(\frac{6.242×{10}^{18}\mathrm{eV}}{1\mathrm{J}}\right)\phantom{\rule{0ex}{0ex}}=0.136\mathrm{eV}$

Hence, the energy of the electron is 0.136 eV . ### Want to see more solutions like these? 