Rutherford's Nuclear Size *

If the positive charge in gold atoms were uniformly distributed over a sphere or radius 5 Angstroms, what is the maximum $\alpha$ particle kinetic energy for which the $\alpha$ can be scattered right back in the direction from which it came?

To solve this, we need to compute the potential at the center of the charge distribution relative to the potential at infinity (which we will say is zero). This tells us directly the kinetic energy in eV needed to plow right through the charge distribution.

The potential at the surface of the nucleus is ${1\over 4\pi\epsilon_0} {Ze\over R}$ where Z is the number of protons in the atom and R is the nuclear radius. That's the easy part. Now we need to integrate our way into the center.

$$\bgroup\color{black}V={1\over 4\pi\epsilon_0} {Ze\over R}-\in... ...R^0{1\over 4\pi\epsilon_0} {r^3\over R^3}{Ze\over r^2}dr\egroup$$

The ${r^3\over R^3}$ gives the fraction of the nuclear charge inside a radius $r$.

$$\begin{eqnarray*} V={1\over 4\pi\epsilon_0} {Ze\over R}-{1\over 4\pi\epsilon_0 ... ...t({Ze\over R}+{Ze\over 2R}\right)={3Ze\over 8\pi\epsilon_0R}\\ \end{eqnarray*}$$

So

$$\bgroup\color{black}V={(3)(79)(1.6\times 10^{-19}\mathrm{C})\... ...{C^2/Nm^2})R} ={1.7\times 10^{-7}\over R} \mathrm{Nm/C}\egroup$$

The is then the kinetic energy in eV needed for a particle of charge $+e$ to plow right through the center of a spherical charge distribution. The $\alpha$ particle actually has charge $+2e$ so we need to multiply by 2. For a nuclear radius of 5 Åor $5\times10^{-10}$ meters, we need about 680 eV to plow through the nucleus. For the actual nuclear radius of about 5 Fermis or $5\times10^{-15}$ meters, we need 68 MeV to plow through.

Lets compare the above SI units numbers to my suggested method of using the fine structure constant... Putting in the alpha charge of $2e$.

$$\bgroup\color{black}U={6Ze^2\over 8\pi\epsilon_0R}={3Z\alpha\... ...)(79)(1973)\over (137)(5)} \mathrm{eV}=683 \mathrm{eV}\egroup$$

This is easier.