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 \bgroup\color{black}$\alpha$\egroup particle kinetic energy for which the \bgroup\color{black}$\alpha$\egroup 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 \bgroup\color{black}${1\over 4\pi\epsilon_0} {Ze\over R}$\egroup 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.

\begin{displaymath}\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\end{displaymath}

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

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}\\


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

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

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

This is easier.

Jim Branson 2013-04-22