I read this link

How did Rutherford conclude that most of the mass (as well as the positive charge) was concentrated in the nucleus?

And understand from it that the scattering pattern is consistent with an alpha particle colliding with a much more massive particle.

Like a billiard ball colliding with a bowling ball, there will be a fraction on billiard balls that bounce back. (unlike two billiard ball collisions where a head on collision never results in a reflection back - unless the impacting billiard ball has a lot of backspin which we ignore in this experiment because alpha particles don’t have backspin)

But why did he conclude that there was only one such particle? And why did he conclude that it was at the center?

I know that sounds stupid and someone will bring up a symmetry argument. But consider that the distance between the origin of the alpha particle and its collision >>> diameter of massive object it is colliding with. So where the massive object really is in its neighborhood is really unknown. And if you don’t know where it is in its neighborhood, who is to say that the next collision did not occur in a different region of the neighborhood? So the massive particle may have had its one ‘orbit’ or random motion? Or that the next collisions occurred with another massive particle or particles entirely?


2 Answers 2


There are a few arguments which are part of the answer.

  1. The assumption of a distributed charge over the atom (Thomson model) is discarded due to the fact that the scattering would only be at very low angles and most of the $\alpha$ particles would pass the thin film almost undisturbed.

  2. To have some large scattering angles one needs a "hard" center. And also heavy since the film could be gold (for example) with many nucleons. It was known that the electrons would be lighter and therefore the center of mass would be mostly in the nucleus, and the reduced mass of the system is just the mass of the $\alpha$ particle and the recoil of the nucleus small.

  3. Then the nucleus would be the smart choice for the origin of the coordinate system to study the process, since it does not move much and it is the center of mass.

  4. There is an assumption of spherical symmetry for the charge distribution around the origin in the nucleus but this is because one can assume the nucleus is point-like. This is tested afterwards; indeed, in the experiment one can ask the minimum distance between the projectile and the nucleus. If the $\alpha$ particle penetrates the nucleus the scattering would also be either non-elastic or involve some other process. This did not happen, the minimum distance (derived with energy considerations) would be consistent with a point-like charge also centered at the center of mass and also therefore point-like.

  5. The other thing is that the hypothesis and calculations were consistent with the data of the ratio of scattered particles at a given angle, atomic number and so on.

This is an example of a hypothesis which leads to a calculation for the cross section and then the contrast with data, which was extremely close to the calculation leaving little room for other assumptions.

  • $\begingroup$ forgive my density, but consider a billiard ball and bowling ball. Imagine the bowling ball on a huge billiard table. Now shoot a billiard ball at it a hundred times. Record the scattering angles. A small percentage of the time, the billiard ball will recoil towards you. Now, Move the bowling a ball a little. Repeat the scattering experiment. The distributions should be almost identical if the balls start very far apart. Repeat 100 times, each time moving the bowling ball a littleThe distribution of rebound angles will be the same as if the bowling ball were stationary. $\endgroup$ Apr 28, 2020 at 19:25
  • $\begingroup$ Your arguments do not address my concern. The 'hard center' is not my issue. That mass(bowling)>>>mass(billiard) is the mathematical representation of 'hard center'. I did not say that the bowling ball had a significant cross section compared to the atom. For a hydrogen atom, the nucleus is the size of a grain of sand if the atom is an average sized room. There is no mathematical proof that a massive grain of sand needs to be at the center of the room if you are shooting a micoscopic grain of sand at it from across the state. $\endgroup$ Apr 28, 2020 at 19:25
  • $\begingroup$ From your point of view at the other end of the state the two atoms have the same scattering behavior. The only thing you can conclude is that the grain IN the atom is way, way , way more massive. You cannot prove it is not moving. You cannot prove there is only one and not five or more. $\endgroup$ Apr 28, 2020 at 19:27
  • $\begingroup$ If I understand the idea of a "wandering" nucleus around the atom is answered when one considers that the center of mass lies in the nucleus. A moving nucleus would demand moving the center of mass and would not conserve angular momentum, it is also difficult to explain without extra external forces. The center of mass, in the absence of other forces is very much at rest (there is some recoil but that can be ignored). $\endgroup$ Apr 29, 2020 at 23:15
  • $\begingroup$ No atom above absolute zero is motionless. Since most of the mass is in the nucleus(nuclei), atomic motion implies nuclear motion. I do not know how you derive a non zero angular momentum from a nucleus having random (Brownian?) motion inside an atom. Angular momentum requires a net unidirectional revolution, not random. $\endgroup$ Apr 30, 2020 at 21:32

It was the large angle scattering that at convinced Rutherford that most of the mass of an atom is contained in the nucleus and the interaction was via the Coulomb force.
In the first set of experiments the incoming alpha particle only interacted with the nucleus via the Coulomb force and never got close enough to actually “touch” the nucleus so Rutherford could only give an upper limit to the size of the nucleus.

Subsequent experiments with lower atomic number nuclei lead to discrepancies between theory and experiment at large angles of scattering ie at closest approach to the nucleus of the alpha particles. Rutherford was interpreted the discrepancy as being due to the alpha touching the nucleus an interacting with it via a different force. Such experiments enabled him to estimate the size of the nucleus.

Much later it was Robert Hofstadter who used electron diffraction to investigate the structure of the nucleus.

  • $\begingroup$ Please see my comment above. I understand how his data would show a small nucleus much more massive than an alpha particle and with a much smaller cross section than the atom. If the protons were scattered a la plum pudding, you would never get any alpha particles bouncing back at you. So he said the massive nucleus was small compared to the atom. But that does not preclude that the nucleus was wandering around the atom. Because no matter where the nucleus is in the atom, the data would be the same. $\endgroup$ Apr 28, 2020 at 21:22
  • $\begingroup$ And if you accept that a wandering nucleus can give the same data, what prevents you from saying that there are several nuclei? $\endgroup$ Apr 28, 2020 at 21:23
  • $\begingroup$ By ‘large angles of scattering’ I interpret that as meaning the ‘billiard ball comes back at you’ after collision with a massive bowling ball $\endgroup$ Apr 28, 2020 at 21:29
  • $\begingroup$ Here is a fun simulator. phet.colorado.edu/sims/html/rutherford-scattering/latest/… $\endgroup$ Apr 28, 2020 at 22:02

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