Question

An electron is contained in the rectangular box of Fig. 39-14, with widths ${\mathit{L}}_{\mathbf{x}}\mathbf{=}\mathbf{800}\mathbf{}\mathbf{pm}$, ${\mathit{L}}_{\mathbf{y}}\mathbf{=}\mathbf{1600}\mathbf{}\mathbf{pm}$, and ${\mathit{L}}_{\mathbf{z}}\mathbf{=}\mathbf{390}\mathbf{}\mathbf{pm}$. What is the electron’s ground-state energy?

Short answer

The electron’s ground state energy is 3.2 eV.

Step-by-step solution

Introduction:

The state of a physical system (such as an atomic nucleus or atom) that has the lowest energy of all possible states. — also called ground level.

An electron is defined as a negatively charged subatomic particle.

The mass of an electron is $\mathbf{9}\mathbf{.}\mathbf{1093837015}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{coulomb}$.

Determine whether an electron is contained in the rectangular box:

The energy of $E_0$of the particle in 1-D box is,

$E_n=\frac{n^2{h}^2}{8m{L}^2}$

Where n=1,2,3.........

Here, h is the Planck’s constant, m is the mass of the particle and L is the length of the box.

Determine the electron’s ground state energy:

The energy $E_{n_x,n_y,n_z}$of the particle (electron) in 3-D box is,

$E_{n_x,n_y,n_z}=\frac{h^2}{8m_e}\left(\frac{n_x^2}{L_x^2}+\frac{n_z^2}{L_z^2}+\frac{n_z^2}{L_z}\right)$

Where,

$$\begin{gathered} n_x=1,2,3....... \\ {n}_y=1,2,3....... \\ {n}_z=1,2,3....... \end{gathered}$$

Here, $m_e$is the mass of the electron,$L_x,L_z,\mathrm{and}{L}_z$are the length of the box along x,y and z directions respectively.

The ground state energy level of the particle in 3-D box is$\left(n_x,n_y,n_z\right)=\left(1,1,1\right)$

The ground state energy of the electron in 3-D box is,

$$\begin{gathered} {\mathrm{E}}_{1,1,1}=\frac{{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{S}\right)}^2}{8\left(9.1\times {10}^{-31}\mathrm{kg}\right)}\left(\frac{1^2}{{\left(800\mathrm{pm}\left({\displaystyle \frac{{10}^{-12}\mathrm{m}}{1\mathrm{pm}}}\right)\right)}^2}+\frac{1^2}{{\left(1600\mathrm{pm}\left({\displaystyle \frac{{10}^{-12}\mathrm{m}}{1\mathrm{pm}}}\right)\right)}^2}+\frac{1^2}{{\left(390\mathrm{pm}\left({\displaystyle \frac{{10}^{-12}\mathrm{m}}{1\mathrm{pm}}}\right)\right)}^2}\right) \\ =\frac{{\left(6.626\times {10}^{-34}\mathrm{J}.\mathrm{S}\right)}^2}{8\left(9.1\times {10}^{-31}\mathrm{kg}\right)}\left(\frac{1^2}{{\left(800\times {10}^{-12}\mathrm{m}\right)}^2}+\frac{1^2}{{\left(1600\times {10}^{-12}\right)}^2}+\frac{1^2}{{\left(390\times {10}^{-12}\mathrm{m}\right)}^2}\right) \\ =5.1429\times {10}^{-19}\mathrm{J}\left(\frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{c}}\right) \\ =3.2\mathrm{eV} \end{gathered}$$

Therefore, electron’s ground state energy is 3.2 eV.