When alpha particles hit the gold nucleus shouldn't they form thallium. More over shouldn't it also ionize the gold or any atom it hits how does it remain unchanged unionized?

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    $\begingroup$ Both the nuclei and $\alpha$ particles are positively charged so repulsion is expected; if the collision energy were utterly vast then it might be possible to fuse the two, but I expect that such energies are not experimentally possible even nowadays. $\endgroup$
    – porphyrin
    Commented Jun 15, 2020 at 7:24
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    $\begingroup$ @porfyrins Available energies are able to break kernels into pieces. Juliet Curie discovered induced radioactivity by alpha particle bombardment of light elements. New super heavy elements are discovered by bombardment of heavy kernels ( like 206Pb) by medium mass/charge kernels. As there is the only stable isotope of gold 197Au, there could be formed traces of 201Tl, that decays with halflife about 3 days by an electron capture to stable 201Hg. But very most of alpha particles would just pass, and from the rest, very most would just get deflected. $\endgroup$
    – Poutnik
    Commented Jun 15, 2020 at 8:31
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    $\begingroup$ @Poutnik, that is interesting, I did'n't know about the Curie experiment. In this case some Tl should be formed by your argument and I greatly overestimated the energy needed because I was thinking of laser fusion experiments. $\endgroup$
    – porphyrin
    Commented Jun 15, 2020 at 9:36
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    $\begingroup$ We don't care if a few gold atoms turned into thallium. Ditto for ionization. This is not what we measure here. $\endgroup$ Commented Jun 15, 2020 at 9:50
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    $\begingroup$ @Poutnik - you need another particle (likely a neutron) departing the compound nucleus, so the likely isotope formed is 200Tl. In fact, the reverse reaction, TL-200(N,A)AU-197, can be found in the Evaluated Nuclear Data files, although one needs a fairly energetic (> 3MeV) neutron to make it happen. $\endgroup$
    – Jon Custer
    Commented Jun 15, 2020 at 13:41

1 Answer 1


The Rutherford formula, as derived, assumes purely elastic scattering from the Coulomb force. No formation of a compound nucleus is considered. Generally, for most of the initial experiments, the available alpha particle energies from various decays (in the range of a few MeV) were not high enough for large deviations from pure Rutherford scattering. Of course, folks rapidly wanted both other energies (or controlled energies) or other charged particles (such as the protons used by Cockroft and Walton for their first accelerator measurements).

These days, much higher energy particles are available, and are commonly used in nuclear physics. A nice recent paper on the topic is in Physics Review C, "Cross section of $\alpha$-induced reactions on $^{197}$Au at sub-Coulomb energies". Here, the authors measure the reaction cross-sections for $^{197}$Au($\alpha$,$\gamma$)$^{201}$Tl, $^{197}$Au($\alpha$,n)$^{200}$Tl, and $^{197}$Au($\alpha$,2n)$^{199}$Tl. Although mainly of interest to astrophysical heavy-element nucleosynthesis, the results bear on this question.

The bottom line is that high (> 13MeV) $\alpha$ energies are needed to drive these reactions. At the lowest energy used (~ 13.7 MeV), the cross-section for the $^{197}$Au($\alpha$,n)$^{200}$Tl reaction is 0.67 $\mu$b (micro barns). At 20 MeV it has increased to 37 mb (milli barns).

Anyway, it is quite clear that these reactions do not occur to any measurable extent in the few MeV regime.


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