91Ó°ÊÓ

The width of the potential well is

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

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

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

The mass of the electron is

$m_e=9.11×{10}^{-31}\mathrm{kg}$

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

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

The energy in electron volt is

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

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

The mass of the proton is

$m_p=1.67×{10}^{-27}\mathrm{kg}$

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

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

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