The Relationship between coulomb collision, inverse square law and Rutherford scattering

Question

I was reviewing the following question.

"A key result of the α particle scattering experiment of Geiger and Marsden was that the number of particles scattered into a given angle was proportional to the thickness of the target. Explain why this shows that the scattering is of single atoms or nuclei,and not due to multiple scattering off many atoms."

The answer was given as We know that N(θ > 90) is proportional to the number of target nuclei and proportional to t, the thickness of the target. If multiple scattering was occurring our path would consist of a series of small angle scatterings due to the Coulomb field of the nucleus. On average the scattering will be small. For multiple interactions, it should be distributed as a Gaussian with mean 0 and width σ ∝ t. Increasing the thickness will only increase the number of small scattering which will not generally add but will also cancel each other. As a result the angular distribution will be a Gaussian with width proportional to √t which cannot explain the Geiger and Marsden observation.

I was fairly satisfied with this answer however there was one particular aspect I didn't understand which is in bold above. Coulomb collision and the inverse square law don't seem to give any specific satisfactory answers as to WHY small angle scatterings occur? So my question is, why does the coulomb force give small angle scattering?

I believe there is a fundamental piece of mathematics I am missing here. Any direction would be much appreciated.

Answer 1

If you consider this phrase (If multiple scattering was occurring our path would consist of a series of small angle scatterings due to the Coulomb field of the nucleus) in isolation, it does not have to be true. It is possible, in principle, that we have multiple scattering, but scattering on a single atom is still large-angle. But in this case we would have a lot of large-angle scattering, which was not the case in the Geiger-Marsden experiments (very few alpha particles were scattered at a large angle).

I would add that the phrase "A key result of the $\alpha$ particle scattering experiment of Geiger and Marsden was that the number of particles scattered into a given angle was proportional to the thickness of the target" does not have to be true either, when considered in isolation. The phrase probably needs some qualification, like that in the answer ($\theta$>90), because for very small scattering angles one probably has multiple scattering.

Maybe you should consider the question and the answer in the context of the conclusions that were eventually derived from the results of the experiments. So Geiger and Marsden observed alpha particles scattered at large angles. One can ask: is this result consistent with the plum pudding model? It can be, but for the plum pudding model an alpha particle is always scattered by an atom by a very small angle, so large scattering angle can only be a result of multiple scattering, which possibility is eliminated by the fact that the number of particles scattered by a large angle is proportional to the thickness of the target.

Answer 2

The nucleus is 5 orders of magnitude smaller than the atom. This means that nuclei are very far apart from each other. If the alpha particle was a football, the atom would have the size of the football pitch and its nucleus would be another football in that pitch. The chance of shooting the first ball in the direction of the pitch and hitting the second ball is definitely too small. As the incident particle and the scattering center are most of the times far from each other, the interaction and thus the deflection are quite small. You can see this from formula for the deflection angle $\theta$ of the alpha particle, $$\tan(\theta/2)=\frac{q_1q_2}{4\pi\epsilon_0 mbv^2},$$ where $b$ is the impact parameter and $v$ is the velocity of the alpha particle. Using the data from wikipedia, $m= 6.64424×10^{−27} kg$, $q_1 = 2×(1.6×10^{−19}) C$, $q_2= 79×(1.6×10^{−19}) C$, $v = 2×10^{7} m/s$, it gives $$\tan(\theta/2)=\frac{1.37\cdot 10^{-14}}{b}.$$ Plugging an impact parameter of order $10^{-5}m$ and you get that $\theta$ is of order $10^{-9}rad$. Note that this is a classical result.