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In this experiment, is it possible that some of the alpha particles are deflected by the electrons? Gold, after all, usually also has ~79 of them in each atom.

Since the alpha particles want electrons, couldn't some of them steal electrons on their way through, creating Helium+ ions (or atoms)?

Update: @pwf found some good quotes below from Rutherford's report that seem to confirm that he assumed the effects of the electrons on the alpha beam would be low, but he did not measure them nor account for them in his measurements. So this question is updated to recognize that there seems to be general agreement that the deflection by electrons would happen, but that the frequency and angle of deflection would be low. One question remains: If Rutherford didn't bother to measure this effect experimentally, has anyone?

Found this detailed "thought experiment" where he seems to lay out all the calculations very well. Perhaps he or others have conducted the actual experiment:

http://www.med.harvard.edu/JPNM/physics/didactics/physics/charged/lect.html

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    $\begingroup$ The incident alphas definitely do interact with electrons - it is the electronic stopping part of the energy loss term. However, kinematics will quickly show you that an alpha cannot back scatter off an electron - it is not physically possible. $\endgroup$ – Jon Custer Mar 7 '15 at 18:28
  • $\begingroup$ @Floris yes, thank you I should have kept it at "...or even Helium ions or atoms..." $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 14:37
  • $\begingroup$ @john custer, thank you. Vladimir below seems to think "Electrons can deflect a heavy alpha-particle backward if they are sufficiently energetic." I guess the key is how "sufficiently energetic" they would need to be. I think you are saying they could "scatter," but just not "back scatter"? Just curious if the effect of the electron scattering could be distinguishable from the nucleus scattering and, if so, if that has even been measured? thanks $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:00
  • $\begingroup$ Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:21
  • $\begingroup$ The amount of additional electrons you can add is minuscule compared to the number of electrons already there. This would be a very hard experiment. Better use a collider - shoot really high energy electrons in one direction and alpha particles in the opposite direction. I suspect it has been done. $\endgroup$ – Floris Mar 8 '15 at 16:19
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Rutherford modeled the atom as an extremely compact positive nucleus surrounded by a uniform ball of negative charge the "size" of the atom. He included the effect of scattering by the electrons under the assumption that they acted like such a diffuse cloud of negative charge, and showed that such a cloud had negligible probability of scattering the alpha particles by more than a degree. That's not because it lacks mass, but because it is diffuse, or rather, composed of many small charges rather than one compact point of high charge, so the alpha particle trajectory would be composed of multiple small deflections ("deflexions" in his spelling!) from the negative charges rather than a single large one. So yes, the electrons could scatter the alpha particles, but for a diffuse electron cloud the effect is negligible. (Actually, as far as I can tell the data would also be consistent with a compact negative nucleus surrounded by a diffuse positive cloud. How Rutherford justified the conclusion that the nucleus was the positive part is mysterious to me.)


Edit: From Rutherford, Philosophical Magazine 21, 669-688 (1911), penultimate paragraph: "The deductions from the theory so far considered are independent of the sign of the central charge, and it has not so far been found possible to obtain definite evidence to determine whether it be positive or negative." He goes on to explain why a positive nucleus might make more sense in terms of $\beta$ absorption and $\alpha$ emission (e.g. why $\alpha$ particles are emitted with such high velocity).


From the same article: Rutherford explains that J. J. Thomson's model of distributed positive and negative charges involves multiple small scattering events, and predicts only small overall "deflexions" (pp 669-670).

After laying out his own model of the compact central charge surrounded by a uniformly dense negative charge, he writes (pp. 671-2), "Since R [the radius of the ball of negative charge] is supposed to be of the order of the radius of the atom, viz. $10^{-8}$ cm., it is obvious that the $\alpha$ particle before being turned back penetrates so close to the central charge, that the field due to the uniform distribution of negative electricity may be neglected. In general, a simple calculation shows that for all deflexions greater than a degree, we may without sensible error suppose the deflexion due to the field of the central charge alone. Possible single deviations due to the negative electricity, if distributed in the form of corpuscles, are not taken into account at this stage of the theory. It will be shown later that its effect is in general small compared with that due to the central field."

He gets around to that discussion near the end, which may be most relevant to your question (p. 686): "In comparing the theory outlined in this paper with the experimental results, it has been supposed that the atom consists of a central charge supposed concentrated at a point, and that the large single deflexions of the $\alpha$ and $\beta$ particles are mainly due to their passage through the strong central field. The effect of the equal and opposite compensating charge supposed distributed uniformly throughout a sphere has been neglected. Some of the evidence in support of these assumptions will now be briefly considered. For concreteness, consider the passage of a high speed $\alpha$ particle through an atom having a positive central charge Ne, and surrounded by a compensating charge of N electrons. Remembering that the mass, momentum, and kinetic energy of the $\alpha$ particle are very large compared with the corresponding values for an electron in rapid motion, it does not seem possible from dynamic considerations that an $\alpha$ particle can be deflected through a large angle by a close approach to an electron, even if the latter be in rapid motion and constrained by strong electrical forces. It seems reasonable to suppose that the chance of single deflexions through a large angle due to this cause, if not zero, must be exceedingly small compared with that due to the central charge."

BTW, I highly recommend reading the original article. Once you decipher the somewhat archaic notation, it's extremely readable.

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  • $\begingroup$ By the time of Rutherford's experiment the cathode rays had been identified as negatively charged and much lower in mass than atoms. Thus the expectation that the positive charge would be associated with the largest part of the atoms mass which was also present in the "plum pudding" model of the atom that Rutherford used in devising his expectations about how the experiment would come out. And the mass of the electrons does matter if they have low momentum (implied in the models of the time and largely true). $\endgroup$ – dmckee Mar 10 '15 at 0:16
  • $\begingroup$ Yes, I agree there was enough evidence to suspect that electrons were low-mass, given e/m measurements and existing estimates of e. But I think Rutherford's analysis and results were consistent with the low-mass, negative charge being the clumped, highly concentrated charge within the atom and the positive (though massive) charge to be the distributed part. The question was only how the charge was distributed in the atom; in terms of inertia, the atom was treated as a unit. (That makes sense assuming the positive and negative charges are reasonably tightly bound together.) (cont'd) $\endgroup$ – pwf Mar 10 '15 at 3:57
  • $\begingroup$ (cont'd) Besides, I don't think it was entirely accepted that the charges in cathode rays (assuming they were discrete particles) were the same as the negative charges in atoms, was it? In any case, I don't think there's any mention of the negative charges' mass in Rutherford's paper, independent of the atom's mass. $\endgroup$ – pwf Mar 10 '15 at 4:03
  • $\begingroup$ @pwf: thank you. Do you have a source for this: "He...showed that such a cloud had negligible probability of scattering the alpha particles by more than a degree."? This seems to be the closest so far in answering the question. It seems he realized the electrons could deflect, but I wonder if the "more than a degree" was a guess, or did he somehow isolate and verify this assumption through testing? $\endgroup$ – Brad Cooper - Purpose Nation Mar 10 '15 at 13:58
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    $\begingroup$ I'll add some quotes from Rutherford (1911) showing how far he considered this question to my answer. $\endgroup$ – pwf Mar 10 '15 at 16:40
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The key difference is that the alpha particle is several thousand times heavier than the electron. It would be like you rolled a bowling ball at a marble and it bounced backward. As Rutherford said:

"It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus. It was then that I had the idea of an atom with a minute massive centre, carrying a charge."

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  • $\begingroup$ Thank you. But even if you rolled a bowling ball and it hit a marble, the marble could still deflect the bowling ball off a straight course. And if the marble stuck to the bowling ball as it rolled, it would put it into an erratic roll as well. It seems like the alpha wants to grab electrons and seems like a hydrogen atom would act differently as it traveled than an alpha particle? $\endgroup$ – Brad Cooper - Purpose Nation Mar 7 '15 at 18:23
  • $\begingroup$ Well, maybe, but remember that the detector probably doesn't give a click or a dot or whatever from a collision with neutral hydrogen. $\endgroup$ – zeldredge Mar 7 '15 at 18:42
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    $\begingroup$ A standard surface barrier detector will register energetic neutrals. $\endgroup$ – Jon Custer Mar 7 '15 at 18:46
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    $\begingroup$ @PurposeNation The small angle deflections due to electrons can not be disentangled from those due to the nuclei. More over, in the original experiment the beam was only roughly columnated, so neither could be easily distinguished from the natural angular spread of the beam. $\endgroup$ – dmckee Mar 7 '15 at 22:39
  • $\begingroup$ @dmckee Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:21
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Electrons can deflect a heavy alpha-particle backward if they are sufficiently energetic. There are such electrons indeed, but their concentration is very small, I guess.

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    $\begingroup$ "Their concentration is very small"? No - there are lots of electrons. The point is that the alpha particle will not be very bothered by the interaction with the electrons - that's just a consequence of their relative mass. This means that it is hard to measure this; but it's quite possible that some alpha particles will come out of the gold foil with an electron (or even two) attached. $\endgroup$ – Floris Mar 7 '15 at 23:19
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    $\begingroup$ @Floris I think he means that electrons with large Fermi momentum back along the beam (which could in fact scatter an incident alpha pretty substantially) are rare, which is a vast understatement when we're talking about momenta comparable to that of the beam alphas. $\endgroup$ – dmckee Mar 8 '15 at 2:15
  • $\begingroup$ @dmckee re-reading the answer I see your point - thanks $\endgroup$ – Floris Mar 8 '15 at 3:21
  • $\begingroup$ If this is true, I wonder if Rutherford (or anyone doing this experiment since) attempted to measure and account for this type of electron deflection vs. deflection by the gold nucleus? has this effect ever been measured? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 14:56
  • $\begingroup$ @Vladimir: Perhaps the effects of the electrons could be measured by adding various amounts of electric charge to the foil? $\endgroup$ – Brad Cooper - Purpose Nation Mar 8 '15 at 15:23
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Note that there is an effect on the electrons that is much larger than caused by direct interaction with the alpha particle. If the alpha partcle collides with the nucleus then the nucleus will change its momentum. To a good approximation this is an instanteneous change and then the so-called "sudden approximation" applies to the electron state. So, the electrons find themselves in a new moving potential well relative to the old potential well. Because this change happens almost instantenously, the wavefunction stays the same. But the old wavefunction is not an energy eigenstate, this leads to a finite probability that the electron configuration will not be the ground state configuration.

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Exept that alpha particles are much heavier than electrons and it wouldn't make sense for them to change the direction when in contact with electrons, the point of the experiment was that, because of the deflected alpha particles ONLY in a very tiny, central area of the atom, most of the atom space is EMPTY. This conclusion wouldn't been made if alpha particles and electrons would somehow interact.

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  • $\begingroup$ Thanks, Ndrina. However, as confirmed by @pwf, Rutherford himself acknowledged that some deflection by electrons would occur (see the references). $\endgroup$ – Brad Cooper - Purpose Nation Mar 15 '15 at 15:27
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Backscattering requires a repulsive force while an electron would attract the alpha particle. Furthermore, the electron is lighter than the alpha particle : in a collision between a truck and a mosquito, the truck doesn't rebound at all.

EDIT : unidimensionnal motion (ie. the particle moving back and forth on the same line) requires a repulsive force, but dmckee argued that an attractive force could produce a result that would be experimentally undistinguishable. To rule out the possibility of scattering by an electron, only the difference in mass in necessary : for a negative charge to backscatter an alpha particle, it would need to be much heavier than an electron. Now, the mass of the electron was known to be very small at the time of Rutherford's experiment. Even if ALL the electrons in the gold atom would gatter in a small nucleus, it would not be enough : by Newton's third law, an equal force would be exerted on the alpha particle and on the negative charge, therefore the latter would have 80 times more acceleration. The negative charge would be scattered, not the alpha particle.

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    $\begingroup$ Notice that the sun's gravity (a strictly attractive force) can scatter comets back toward their original direction of approach... $\endgroup$ – dmckee Mar 7 '15 at 22:36
  • $\begingroup$ actually no, it cannot. The only way something is scattered at 180 degrees is with a repulsive force. $\endgroup$ – Teacher77 Mar 9 '15 at 0:38
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    $\begingroup$ Consider an object on a parabolic orbit ($E = 0$) at time long before and long after the encounter. It's final momentum is essentially anti-parallel to it's initial momentum. Allow "long time" to grow without bound and it is exactly anti-parallel. $\endgroup$ – dmckee Mar 9 '15 at 0:46
  • $\begingroup$ In that limit, the incident direction and the returning direction would be two parallel directions with an infinite distance between the two. NOT the same direction. Perfect back scattering (ie. one dimensional motion where the particle simply inverts its motion) can only be acheived with a repulsive force. P.S. A parabolic orbit has e = 1, not 0... $\endgroup$ – Teacher77 Mar 9 '15 at 3:58
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    $\begingroup$ Compute the angle of each track. They approach the same value because the in-and-out coordinate grows faster than the width coordinate. And in particle physics the "solar system" is less than a nanometer across, so the detector a few cm away is a good approximation to infinite remove. $\endgroup$ – dmckee Mar 9 '15 at 4:04

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