A hydrogen atom (with the Bohr radius of half an angstrom) is situated

between two metal plates 1 mm apart, which are connected to opposite terminals of a 500 V battery. What fraction of the atomic radius does the separation distance d amount to, roughly? Estimate the voltage you would need with this apparatus to ionize the atom. [Use the value of in Table 4.1. Moral:The displacements we're talking about are minute,even on an atomic scale.]

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

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

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

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

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

The expression for electric field is:

$E=V/x$

Substitute the values of in the above equation and get

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

The expression for the dipole moment is,

$$\begin{gathered} p=\alpha E \\ =ed \end{gathered}$$

Thus,

$d=\frac{\alpha E}{e}$

Substitute the values in the above equation and get

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

The ratio of the separation distance to the atomic radius is

$r=d/R$

Substitute the values in the above equation and get

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

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

The expression for the voltage required to ionize the atom is

$V_i=\frac{Rex}{\alpha }$

Substitute the values in the above equation and get

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

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