Talk:Rutherford backscattering spectroscopy

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Geiger Marsden diagram - plum pudding misleading Hi, I think the first part of the diagram is misleading. If the positively charged nuclear particles were evenly distributed throughout the plum pudding then that model predicts there would be a "significant" proportion of the "projected" alpha particles deflected from a straight path. So on the receiving surface one would see light flashes scattered over a "significant" area - the angle of deflection dependent on how close the alpha particles came to the positively charged nuclear particles (I can't remember if protons were named at that time or if neutrons had been proposed). Anyway, what Rutherford's assistants saw was that most of the light flashes were on a path that indicated there was no deflection of the alpha particles by the atoms. Hence the deduction was the gold(?) atoms were mainly "empty space". The clincher for a mainly empty space but with concentrated nucleus was the detection of a few "bounced back" alpha particles. Hence the interpretation of the highly concentrated positive nucleus. Most people when they tell the story concentrate on the "bounced back" "shell at tissue paper" part of the interpretation , but the lack of deflection of the alpha particles is just as significant in providing evidence of the nuclear model. Now ..... my point about the entry is .... the first part of the diagram (plum pudding) should show a "significant" proportion of the paths of the alphas particles being deflected by the positive particles in the nucleus - not passing straight through as in the current version of the picture. Apart from that, I like the way the diagram is drawn and put together the compare and contras the plum pudding and nuclear models. Clive long 07:58, 17 October 2007 (UTC) Clive Long

A fact from Rutherford backscattering spectroscopy appeared on Wikipedia's Main Page in the Did you know? column on 17 October 2007.
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[edit] Factor missing

The differential cross-section of the backscattering is missing the prefactor \frac{1}{4\pi \epsilon_0} .. and don't tell me it's in atomic units, as there would be no e2 then. --MarsmanRom —Preceding unsigned comment added by 141.20.43.124 (talk) 09:55, 7 February 2008 (UTC)