91Ó°ÊÓ

The distance between metal plates is: $\mathrm{x}=1\mathrm{mm}=\left(1\mathrm{mm}×\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=0.001\mathrm{m}$.

The emf of the battery is: $V=500V$.

The Fermi constant is:$\mathbf{Î\pm }\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{32}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{41}}\mathbf{}\mathbf{C}\mathbf{×}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{V}$ .

The electric charge is: $\mathbf{e}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{C}$.

The atomic radius is: $\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{}\mathbf{m}$.

The expression for electric field is:

$E=V/x$

Substitute the values of in the above equation and get

$$\begin{gathered} E=500V/0.001m \\ =5×{10}^{5}V/m \end{gathered}$$

The expression for the dipole moment is,

$$\begin{gathered} p=\mathrm{Î\pm }E \\ =ed \end{gathered}$$

Thus,

$d=\frac{\mathrm{Î\pm }E}{e}$

Substitute the values in the above equation and get

$$\begin{gathered} d=\frac{7.34×{10}^{-41}\mathrm{C}.{\mathrm{m}}^2/\mathrm{V}×5×{10}^5\mathrm{V}/\mathrm{m}}{1.6×{10}^{-19}\mathrm{C}} \\ =2.2×{10}^{-16}\mathrm{m} \end{gathered}$$

The ratio of the separation distance to the atomic radius is

$r=d/R$

Substitute the values in the above equation and get

$r=\frac{2.2×{10}^{-16}\mathrm{m}}{0.5×{10}^{-10}\mathrm{m}}$

Thus, the separation distance is $4.6×{10}^{-6}$ times the atomic radius.

The expression for the voltage required to ionize the atom is

$V_i=\frac{Rex}{\mathrm{Î\pm }}$

Substitute the values in the above equation and get

$$\begin{gathered} V_i=\frac{0.5×{10}^{-10}\mathrm{m}×1.6×{10}^{-19}\mathrm{C}×0.001\mathrm{m}}{7.32×{10}^{-41}\mathrm{C}.{\mathrm{m}}^2/\mathrm{V}} \\ ={10}^8\mathrm{V} \end{gathered}$$

Thus, the voltage required to ionize the atom is ${10}^8\mathrm{V}$.