Re: How does Rutherford measure the size of a nucleus from his formula?

Hello Abdullah,

There are many interesting things to learn from Rutherford scattering---you hint at two different ones in your question.

First of all, consider an experiment where the alpha particles are *not* energetic enough to actually hit the nucleus. The alphas are scattered by the electrostatic (Coulomb) repulsion of the two positively-charged objects. You can measure the probability for an alpha particle to be deflected at 10 degrees, 20 degrees, and so on up to 180 degrees ("backscattering"). Then you ask, "Are these probabilities are the same as I expect from pure-Coulomb repulsion? I.e., do the probabilities match the Rutherford formula?" If the scattering indeed looks like Rutherford scattering, you have learned something: you can put an upper limit on the size of the nucleus. This is how to do it: You know that the energy of your alpha particle starts off as E0. You know that, in order to bring the alpha particle (charge z=2e, where e is the electron's charge) to within a distance R of the nucleus (charge Z = atomic number * e), it needs to pile up electrostatic potential energy EP = (Z z)/(4 pi epsilon_0 R^2). The sharply-backscattered particles are the ones that got as close to the nucleus as possible; they must have had EP=E0 at their closest approach to the nucleus. If these alphas still obey the Rutherford formula, you can infer that they only felt point-source-Coulomb forces on their very close approach, and therefore the nucleus must be smaller than R. That's what Rutherford did; his very first data with the gold foil clearly shows that the "Plum Pudding" model of the atom is incorrect.

OK, what happens when you do high-energy Rutherford scattering, and the alpha particles *do* crash into the nucleus? What else do you learn? That's very complicated, actually. It was simple in the Plum Pudding model: for this model, imagine the nucleus is a uniform sphere of charges, held together for no good reason in a sphere (say) 0.1 nanometers across. There's no "strong force"; you expect the alpha particle to Coulomb-scatter off of these charges. In this case, once the alpha particle gets "inside" the 0.1-nm sphere, it feels a much weaker repulsion---since there is charge on all sides, in front and behind---and it is not pushed back as forcefully as it would if all the charge were still in front of it. Therefore, you would not see any "hard scattered" alphas---reflecting back in the direction of the beam---if the alpha can get within 0.1 nm of the atom.

Once we add the strong force, things get complicated. Now, we can't use a simple criterion like "After the alpha crosses radius R, it feels less force pushing backwards." The alpha particle can interact with the nucleus in many ways: resonant scattering, for example, in which the alpha and nucleus momentarily form a "bound state" which then releases the alpha in a random direction. You can get nuclear reactions---the alpha can get absorbed into the nucleus, or can knock out a neutron, or can kick the nucleus into emitting a show of gamma rays. In practice, figuring out the size and shape of the nucleus involves looking at *all* of these different reactions, and fitting them to different nuclear models. In general, though, I think that alpha-nucleus scattering, for alpha energies which are greater than the Coulomb barrier but not so high that they reliably break up the nucleus, will show an excess of backscattering when compared to the Rutherford formula.

I hope this helps! There's a nice page on the Rutherford formula at Georgia State's "Hyperphysics" site. If you're interested in digging into the nitty-gritty of nuclear scattering and interactions, and you're willing to learn how to read some ugly text files, poke around the National Nuclear Data Center, where you can download experimental data on all sorts of nuclear reactions. Let me know (via MadSci) if you want something specific and need help finding it.

-Ben