What is the ground-state energy of

(a) an electron and

(b) a proton

if each is trapped in a one-dimensional infinite potential well that is 200 wide?

The width of the potential well is,

$$\begin{gathered} L=200\mathrm{pm} \\ =200\times \left(1\mathrm{pm}\times \frac{1\mathrm{m}}{{10}^{12}\mathrm{pm}}\right) \\ =2\times {10}^{-10}\mathrm{m} \end{gathered}$$

The ground state energy of a particle of mass min an infinite potential well of widthLis

${\mathit{E}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}\mathbf{m}{\mathbf{L}}^{\mathbf{2}}}$ ..... (1)

Here, h is the Planck's constant having value

$\mathit{h}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{‐}\mathbf{34}}\mathbf{}\mathbf{J}\mathbf{.}\mathbf{s}$

The mass of the electron is

$m_e=9.1\times {10}^{-31}\mathrm{kg}$

From equation (1) the ground state energy of electron is

$$\begin{gathered} E_0=\frac{{\left(6.6\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times 9.1\times {10}^{-31}\mathrm{kg}\times {\left(2\times {10}^{-10}\mathrm{m}\right)}^2} \\ =15\times {10}^{-19}\times \left(1\mathrm{J}\right)·\left(1\mathrm{J}\times \frac{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)·\left(1{\mathrm{s}}^2\right)·\left(\frac{1}{1\mathrm{kg}}\right)·\left(\frac{1}{1{\mathrm{m}}^2}\right) \\ =15\times {10}^{-19}\mathrm{J} \end{gathered}$$

The energy in electron volt is

$$\begin{gathered} E_0=15\times {10}^{-19}\times \left(1\mathrm{J}\times \frac{0.624\times {10}^{19}\mathrm{eV}}{1\mathrm{J}}\right) \\ =9.36\mathrm{eV} \end{gathered}$$

The required energy is 9.36 eV .

The mass of the proton is

$m_p=1.67\times {10}^{-27}\mathrm{kg}$

From equation (I) the ground state energy of proton is

$$\begin{gathered} E_0=\frac{{\left(6.6\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times 1.67\times {10}^{-27}\mathrm{kg}\times {\left(2\times {10}^{-10}\mathrm{m}\right)}^2} \\ =0.0082\times {10}^{-19}\times \left(1\mathrm{J}\right)·\left(1\mathrm{J}\times \frac{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)·\left(1{\mathrm{s}}^2\right)·\left(\frac{1}{1\mathrm{kg}}\right)·\left(\frac{1}{1{\mathrm{m}}^2}\right) \\ =0.0082\times {10}^{-19}\mathrm{J} \end{gathered}$$

The energy in electron volt is

role="math" localid="1661763664410" $$\begin{gathered} E_0=0.0082\times {10}^{-19}\times \left(1\mathrm{J}\times \frac{0.624\times {10}^{19}\mathrm{eV}}{1\mathrm{J}}\right) \\ =0.005\mathrm{eV} \end{gathered}$$

Hence, the required energy is $0.005\mathrm{eV}$.