A reader of Ernest Rutherford’s famous paper “The Scattering of α and β Particles by Matter and the Structure of the Atom,” which reports the discovery of the atomic nucleus, may be perplexed by a couple of expressions at the beginning of §3. An explanation follows.
We are playing a game of shooting tiny particles at a thin screen of large spherical atoms, in the direction normal (perpendicular) to its surface. With very few exceptions, the particles pass straight through the screen; for our present purpose we assume that all do. Having chosen a distance d, we say that a particle hits an atom if it comes within that distance of the center of the atom. For a given shot, our score is the number of atoms hit by the particle. Each shot is taken in the same way, and the shots are independent of one another—the score for one shot has no influence on the score for another.
When we take a number of shots, our average score means the result of dividing the total of our scores by the number of shots. For example, if we score successively 2, 0, 5, 2, 0, 2, 3, our total score is 14 and the number of shots is 7, so our average score is 2. If we score successively 1, 0, 0, 0, 1, 0, 1, 0, our total score is 3 and the number of shots is 8, so our average score is 0.375.
Since the shots are alike and independent, it is reasonable to suppose that as we take more and more of them, our average score will approach some definite value as a limit. The limiting value can be called our expected score, or expectation, because when we play the game, we expect our scores to cluster around that value, and our average score to tend towards it. The expectation, which we call E, will be well approximated by the average score we get when we take a large number of shots.
In order to estimate the expectation, let us call the thickness of the screen t, and the density of atoms within it n. The latter quantity means the average number of atoms per unit volume; that is the same as the average number of centers of atoms per unit volume.
The question, On average, how many atoms does a particle hit? is equivalent to the question, On average, how many atom-centers lie within distance d of the path of the particle? and therefore also to the question, On average, how many atom-centers lie within the right circular cylinder of radius d whose axis is the path of the particle through the screen, and whose height is the thickness t of the screen?
The volume of such a cylinder—that is, the number of unit volumes it contains—is the area of its base, which is π d 2, times its height t, or π d 2 t. It follows that the average number of atom-centers in such a cylinder is n × π d 2 t = π d 2 n t. This is then our expectation: E = π d 2 n t.
We now apply this result to two special cases.
Case 1. Let d be the radius R of an atom. In this case, hitting an atom has the ordinary sense of coming into contact with it. The expectation E is π R 2 n t. This is what Rutherford means by saying that “the number of collisions of the particle with the atom of radius R is π R 2 n t.” The value of E is of the order of the number of atoms spanning the thickness t—a few thousand, for a gold-foil screen.
Case 2. Let d be a distance p much smaller than R, a thousand or ten thousand times smaller. In this case it is very difficult to hit an atom—so difficult, that we may assume that no particle hits more than one. Then for each shot our score is 0 or 1, and the expectation E = π p 2 n t has a value between 0 and 1.
Since in this case there are only two possibilities for each shot, either it hits or it doesn’t, we can speak of the probability P that a given shot will hit an atom, understanding by this the proportion of hits, or average score, obtained when a great many shots are taken (more precisely, its limiting value as the number of shots increases)—that is, the expectation: P = E. If we assert, for example, that P = 0.20, or 20%, we expect that out of, say, 1000 shots, there will be about 200 hits—that the average score for those 1000 shots will be about 200/1000 = 0.20, that about 20% of them will be hits. Hence Rutherford writes, “The probability … of entering an atom within distance p of its centre is given by … π p 2 n t.”
Remarks. 1. The assumption in the first paragraph is justified as follows. A particle deviates appreciably from a straight path only if it hits an atom for very small d, as in Case 2. Even if so deflected, it is very unlikely to hit another atom, because d is much less than R and t is very small. Thus only the possibility of a first encounter along a straight path need be admitted.
2. One may wonder why probability enters the picture only with Case 2. Shouldn’t the general case somehow involve probability? It does. Let N be the greatest number of atoms the particle can possibly hit. Let P0 be the probability that it hits no atoms, P1 the probability that it hits exactly one, P2 the probability that it hits exactly two, …, PN the probability that it hits exactly N. Then P0 + P1 + P2 + ∙ ∙ ∙ + PN = 1 and P0 ∙ 0 + P1 ∙ 1 + P2 ∙ 2 + · · · + PN ∙ N = E. In Case 2, the only non-zero probabilities are P0 and P1, hence P1 = E.
3. The intuitive understanding of expectation as the limit of the average score rests on the assumption stated in the third paragraph, that the limit exists. In the mathematical theory of probability a different approach is taken, and the limit assertion becomes a theorem, or rather a class of theorems (laws of large numbers).