Question

An electron is contained in the rectangular corral of Fig. 39-13, with widths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}\mathbf{800}\mathbf{}\mathbf{pm}$ and${\mathbf{L}}_{\mathbf{y}}\mathbf{=}\mathbf{1600}\mathbf{}\mathbf{pm}$. What is the electron’s ground-state energy?

Short answer

The ground state energy of the electron is 0.7375 eV.

Step-by-step solution

Introduction:

The electron is a subatomic particle whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family and are generally thought to be elementary particles because they have no known components or substructure.

Solution:

The energy of the electron in a rectangular corral is.

$E_n=\frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right)$

Here, the Planck constant is $\mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$and the mass of the electron is $\mathrm{m}=9.1\times {10}^{-31}\mathrm{kg}$.

Here the widths is,

$$\begin{gathered} {\mathrm{L}}_{\mathrm{x}}=800\mathrm{pm} \\ =800\times {10}^{-12}\mathrm{m} \\ {\mathrm{L}}_{\mathrm{x}}=1600\mathrm{pm} \\ =1600\times {10}^{-12}\mathrm{m} \end{gathered}$$

For ground state$n_x=1,n_y=1$.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}=\frac{{\left(6.63\times {10}^{-34}\mathrm{Js}\right)}^2}{8\left(9.1\times {10}^{-31}\mathrm{kg}\right)}\left(\frac{{\left(1\right)}^2}{\left(800\times {10}^{-12}{\mathrm{m}}^2\right)}\right) \\ =0.6038\left(\frac{1}{64}+\frac{1}{256}\right)\times {10}^{-17}\mathrm{J} \\ =0.0118\times {10}^{-19}\mathrm{J} \\ =1.18\times {10}^{-19}\mathrm{J} \\ {\mathrm{E}}_{\mathrm{n}}=\frac{\left(1.18\times {10}^{-19}\mathrm{J}\right)}{\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)} \\ =0.7375\mathrm{eV} \end{gathered}$$

Hence, the electrons ground state energy is 0.7375 eV.