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

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

The ground state energy of a particle of mass m in an infinite potential well of width L is

$E_0=\left(\frac{h^2}{8m_e{L}^2}\right)n^2............\left(1\right)$

Here h is the Planck's constant having value

$h=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$

The mass of the electron is

$m_e=9.11\times {10}^{-31}\mathrm{kg}$

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

$$\begin{gathered} E_0=\left(\frac{{\left(6.6\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)}^2}{8\times 9.11\times {10}^{-31}\mathrm{kg}\times {\left(200\times {10}^{-12}\mathrm{m}\right)}^2}\right){\left(1\right)}^2 \\ =15\times {10}^{-19}\times \left(1\mathrm{J}\right)·\left(1\mathrm{J}\times \frac{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)·\left(1{\mathrm{s}}^2\right)·\left(\frac{1}{1\mathrm{kg}}\right)·\left(\frac{1}{1{\mathrm{m}}^2}\right) \\ =15\times {10}^{-19}\mathrm{J} \end{gathered}$$

The energy in electron volt is

$$\begin{gathered} E_0=15.1\times {10}^{-19}\times \left(1\mathrm{J}\times \frac{0.624\times {10}^{19}\mathrm{eV}}{1\mathrm{J}}\right) \\ =9.42\mathrm{eV} \end{gathered}$$

The required energy is $9.42\mathrm{eV}$.

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

$E_1=\left(\frac{h^2}{8m_p{L}^2}\right)n^2..........\left(1\right)$

$$\begin{gathered} E_1=\left(\frac{{\left(6.63\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{8\times 1.67\times {10}^{-27}\mathrm{kg}\times {\left(200\times {10}^{-2}\mathrm{m}\right)}^2}\right){\left(1\right)}^2 \\ =8.225\times {10}^{-22}\times \left(1\mathrm{J}\right)·\left(1\mathrm{J}\times \frac{1\mathrm{kg}·{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)·\left(1{\mathrm{s}}^2\right)·\left(\frac{1}{1\mathrm{kg}}\right)·\left(\frac{1}{1{\mathrm{m}}^2}\right) \\ =8.225\times {10}^{-22}\mathrm{J} \\ =8.225\times {10}^{-22}\times \left(1\mathrm{J}\times \frac{0.624\times {10}^{19}\mathrm{eV}}{1\mathrm{J}}\right) \\ =5.13\times {10}^{-3}\mathrm{eV} \end{gathered}$$