Talk:Plum pudding model

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[edit] Textbook Contradiction

This article mentions a "little known fact" that the electrons in the model were dynamic and stabilised in orbits "by the fact that when an electron moved farther from the center of the positive cloud, it felt a larger net positive inward force, because there was more material of opposite charge, inside its orbit" However, I have a textbook that states "each electron would take up a position in the jelly as far as possible from all other electrons owing to electrostatic repulsion" Can anyone verify the correct model? Textbook source - "Year 12 Senior Physics" Moyle, Allan, Millar, Molde. published 1989. pg 230 Paper.plane.pilot 19 Oct 06

Probably a case of textbook oversimplification. I'd be more inclined to trust histories of science than textbooks, which of necessity must present concepts in an idealised and simplified form. EdC 20:51, 19 October 2006 (UTC)

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Apparantly, the plum pudding model was actually thought up the the anchient greeks, or somthing like that, about 2000-3000 yeras ago. Though they didnt call it the plum pudding, they just called it a mass of positive charge with minute negetive charge to balance it out...so with this in mind, where the hell did the neutrons come from???

The ancient Greeks did not know "positive" from "negative". These terms are from Benjamin Franklin.

Up to 1932, it was thought the nucleus of atoms heavier than hydrogen had nuclear electrons, since the nuclear mass of these atoms was larger than it should have been for the charge (as compared with hydrogen) by about double. But in 1932 the neutron was discovered as a separate particle about as heavy as the proton with no charge, so it replaced the "proton plus nuclear electron." The earlier idea wasn't such a bad model, BTW, since neutrons do decay to protons, electrons (and antineutrinos). But long before 1932 it had been realized that electrons cannot be confined in the nucleus before they are created there as betas, as this would require far too much energy for quantum mechanical reasons. Far more than is available. So the neutron was badly needed. It has no electron "inside" it. Sbharris 19:36, 5 June 2006 (UTC)

[edit] Modern models

The final irony of the Physics is that the pudding model is always more correct than the (classical) notion of a pointlike nucleus. Indeed, the nuclei bound in a solid move around their equilibrium positions, and their positive charges are smeared over a rather wide region. Even in an isolated atom the nucleus, bound only with light electrons, turns around of atomic center of inertia, and that smears its positive charge.

The positive “cloud” in an atom is described with the second atomic form-factor f_nn(q) [1] which strongly depends on the atomic state |n,l,m>. This form-factor stands at the Rutherford elastic cross section, so the purely elastic “backward” scattering is suppressed by |f_nn(q)|². The higher is the initial (and the final for elastic processes) target atom state n, the stronger is suppression of the deflected “backward” projectiles. For example, in the excited Hydrogen atom with n=43 (Rydberg state), the positive cloud is of the Bohr radius. In the ground Hydrogen state the charge is smeared within 30•10^-13 cm, that is certainly larger than the “proper” proton radius.

In a “condensed matter” the positive cloud localization is of the atomic size. So the condensed matter structure resembles the pudding model.

Apart from elastic, there are inelastic second form-factors f_nn’(q). They give the amplitudes of atom exciting due to transmitting the big momentum q to the nucleus at the “backward” scattering.

Ernest Rutherford, his colleagues, and the followers did not resolve the alfa-particle energies with the accuracy of 10-100 eV, so they all measured inclusive cross sections (elastic plus all energetically admitted inelastic) rather than the elastic one.

The inclusive cross section, according to the quantum mechanics, coincides with the Rutherford formula [1]. This means that in calculations of the number of scattered backward projectiles one can safely use the notion of the pointlike (“free”) nucleus, but one should never think that the target atoms (or more generally – the target energy levels) do not get excited.

All this is quite natural. As soon as we agree that the Rutherford formula is the inclusive result, we have to recognize that there is no ground for the pointlike nucleus notion. Piling up different events does not create an objective notion, but the crude classical one.

In other words, the classical “image of pointlike free” nucleus is always the sum of quantum mechanical images of different “pale photographs” of the bound system undergoing all possible transitions in course of “observations” (scattering). Thus, the inclusive picture is literally a cinematographic illusion obtained with superposing all particular images of quite different elastic and inelastic events (frames of dσ_nn'(q)).

By the way, this result is completely opposite to the theories of hidden variables where the “randomness” of quantum mechanics is explained with averaging a deterministic theory trajectories over some hidden parameters.

In practice there is no possibility to distinguish the fast scattered projectiles with precision of about 10 – 20 or 100 eV. It is even not possible to prepare the incident beam with that energy accuracy. That is why dealing only with scattered projectiles gives inevitably the inclusive cross section.

Another matter is observing recoil atoms. The excited atoms radiate. The atoms excited due to hitting electrons (described with the usual atomic form-factor F_nn'(q) under small angle scattering) radiate standard spectral lines. The target atoms excited due to hitting the nucleus (determined with f_nn’(q) under large angle scattering) receive big momenta; therefore their spectral lines will be essentially shifted (Doppler effect). Registering simultaneously the scattered “backward” projectile and the shifted spectral lines permits distinguishing different inelastic processes. Thus, it is possible in principle to measure the elastic and inelastic cross sections separately. The target atoms should obviously be in a gas state of small density in order not to damp the excitations by the interatomic collisions.

Vladimir Kalitvianski.

1. Attenuation of the Rutherford scattering and atom exciting by fast charged particles for large-angle scattering. Ukrainian Journal of Physics, V. 38, N 6, 1993, pp. 851-854, and Preprint of Sukhumi Institute of Physics and Technology 90-8, 1990, V. Kalitvianski, (in Russian).

Vladimir, I am laboring to understand you. The gold nucleus is far heavier than the H nucleus, so charge smearing due to center of mass smearing is far smaller, especially as this is a system of many electrons in counterbalance, not one. And gold atoms vibrate, but at the speed of an alpha particle, they may as well be standing still, and that is why Rutherford's formula gives point-like results. I don't understand what "excited states" you're talking about-- nuclear excited states? The alphas never penetrate the nuclear charge in gold except for an extremely rare tunnelling event (too rare to ever see), so they never reach the nucleus to excite it. If they did, you'd get a big departure from the Rutherford formula for small impact parameters, showing you that the alphas had enough energy to actually experience inelastic scattering. But that isn't seen.

That it's possible to see departures of this type due to inelastic alpha impacts with lower Z nuclei was proven by Rutherford himself, who noted departure from the elastic model at very small impact parameters when scattering alphas from hydrogen. His minimal r corresponded to about 3.5 fm, which is just about the combined radii of an alpha and a proton. This is semi-classical still. So I suppose I don't get what you're trying to say. SBHarris 21:35, 8 August 2006 (UTC)

Dear Steve,

Concerning excited states – it is atomic excited states, not the nucleus ones. Here I consider an atom as a bound state of a “pointlike” nucleus and the electrons. To excite atomic levels you can “hit” the electrons or the nucleus. “Backward” scattering corresponds to “hitting” the nucleus, so the motion of atomic electrons relative to the nucleus gets perturbed. (“Hitting” means, of course, the potential interaction of the fast charged projectile with the nucleus electrostatic potential.) Textbooks on atomic physics replace the nucleus coordinates with the atomic center of inertia coordinates (unnecessary and erroneous simplification), so they “hit” the atomic center of inertia. Doing so, you will never excite the atom however big is the transferred momentum. With this (unnecessary and erroneous) simplification a finny story is connected. In the thirties (1939) A. Migdal tried to calculate the probability of atom exciting with neutron. The neutron interacts only with the nucleus, but if you use the center of inertia variables (rather than the nucleus ones) you cannot excite the atom – it receives the transferred momentum as a whole. On the other hand, it is physically clear that the atom should get excited if one hits the nucleus. A. Migdal spent a lot of efforts trying to calculate this probability, and the solution clue had finally come to him in his sleep. He did manage to get the correct answer by means of considering the problem in the moving reference frame and using then the theory of sudden perturbations.

In reality this probability is obtained immediately and automatically in the first Born approximation if one uses the true nucleus coordinates rather than coordinates of atomic CI in a short-range neutron-nucleus potential [1]. Here I just do not simplify the coordinates of the interacting particles. Nothing else.

You are right saying that the gold nucleus is very heavy, so for the atom of gold the second elastic atomic form-factor is close to unit (do not confuse it with the nuclear form-factor discovered by Hofstadter). Yet, it is possible to “feel” it with using a heavier or a faster projectile. In addition, you can increase the atomic positive cloud size if you prepare the separate target atoms in metastable excited states (Rydberg atomic states n >> 1).

But in a solid state the atomic form-factor is replaced with the “lattice” form-factor that manifests much larger nucleus delocalization. Here the nuclei are “bound” to each other rather than to the light electrons, that is understandable even for a two-atom molecule.

It is also true that the projectile velocity may be much higher than the nucleus velocity in the target. The faster projectile is, the better is the first Born approximation in the quantum mechanics. And the latter gives a nucleus picture (photo) averaged over different “semi-classical” positions of the nucleus bound in the target. This is the origin of the second elastic atomic (or target) form-factor. (The first form-factor describes the negative target cloud.) The system of many electrons does not counterbalance the nucleus motion in the sense that it does not put the nucleus in the center of inertia. There is always a (symmetric or asymmetric) positive cloud, and its size is at least of order (m_e/m_A)a_n where a_n is the target atom size.

The projectile momentum after an elastic collision p’ is numerically very close to that after inelastic collision: p’=p[1-(E_n’ -E_n)/(p²/2m)]^1/2 if the projectile kinetic energy p²/2m is much higher than the atomic energy differences E_n’ -E_n. So it is practically impossible to distinguish (resolve) the scattered particles on the lost energy with that precision. To detect the excitations, it is necessary to deal with the target final states rather than with the projectiles. One of the simplest ways is to use a small density gas target (with separate atoms). The Doppler shift of recoil atoms gives the quantitative description of the scattering from the bound nucleus.

You may say: “Let us get rid of all electrons and other environment. Let us consider the free nucleus as a target. Then it will be pointlike.”

By saying “pointlike” you mean that there is nothing in the space but the sizeless point. You approach the nucleus closer and closer, and there is still nothing in the space. Until you touch the point. What is wrong with this idea?

The first strangeness is that the pointlike particles need long-distance potentials to interact. This need means nothing but some rather extended nature of charged “particles”. The second problem is the stability of charged particle systems. We need to replace the classical trajectories with the De Broglie waves in order to obtain stability. The De Broglie waves mean that the whole space participates in creating a “particle” as a wave packet. The more exact description we elaborate, the father from pointlikeness we go.

Let us look at the Moon. We may describe it as a pointlike mass furnished with the Newtonian potential. In the classical mechanics its motion is potential and elastic. But we observe the Moon with help of light scattering, viz. with help of inelastic processes. The Moon is more complicated than a simple point R(t). It undergoes internal changes all the time, but we attribute its different images to one and the sole body and create the notion of the Moon. So the classical Moon is in fact the sum of all different images (inclusive picture). And in the space between the Moon and the Earth there are we, there is life. Our energy levels are much smaller than the Moon kinetic energy, so our fuss does not change much the Moon orbit, that is why we are tempted to neglect it in the theory. As I said before, in experiment we do not neglect but sum up different low energy events. An exact theory should take this fact into account.

The same is valid for the atomic nucleus as well as for the other “elementary” free particles. In particular, scattering from the free nucleus is not elastic because it is always accompanied with soft (low energy) radiation. I would say that there is always a “low energy life” in the space, so the space with a pointlike nucleus is not actually empty but necessarily “filled”. The pointlike particle cannot be "alone"! Here also, only the sum of all different radiation cross sections gives the “mechanical” cross section (the Rutherford or another elastic one).

As soon as it is so, is it correct to insist on pointlikeness as on the fundamental feature of anything? I do not think so.

Sincerely,

Vladimir Kalitvianski.