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It is well understood that $\alpha$ decay is explained by quantum tunnelling of $\alpha$ particles through the strong potential well of the nucleus.

This set me to wondering about the Rutherford-Geiger-Marsden experiment and wondering how much velocity an $\alpha$ particle would have needed to overcome the coulomb potential barrier of the gold foil in order for the effects of the strong force to be noticed.

A naive calculation neglecting the effects of special relativity yield a value for the velocity of an $\alpha$ particle of approximately $0.25c$.

Given this value and the typical energy of $\alpha$ particles yielding a velocity of about 5% of $c$, the conclusion is that Geiger and Marsden could not have seen effects of the strong force in their experiment without producing higher energy $\alpha$ particles from a particle accelerator.

Given that this naive calculation yields a speed that is an appreciable fraction of the speed of light I redid the calculation taking special special relativity into account.

This yielded a value of the speed of the $\alpha$ particle of $0.999999973722160c$, only some $8 \,{\rm m s^{-1}}$ slower that the speed of light.

Given that the LHC accelerates protons to within about $3 \,{\rm m s^{-1}}$ of the speed of light one could say that it was way beyond the technology of the day.

Except of course this is itself a naive calculation given that quantum effects have not been taken into consideration. One would assume that the velocity could be a little less and the $\alpha$ particle could tunnel into the nuclear well and the effects be observed in the scattering cross-section.

Might a further calculation involving the de Broglie wavelength of a fast alpha particle shed some light on the matter, or is this also too naive an endeavour?

Incidentally have there been any experiments with positively charged particles that show scattering effects of the strong force or are effects due to scattering by the strong force strictly confined to the realms of neutron scattering experiments?

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The radius of a gold nucleus is around 8.1 fm, and if we take this as the distance of closest approach we get the potential energy:

$$ V = \frac{\kappa Q_1 Q_2}{r} \approx 4.5 \,\text{pJ} $$

The relativistic expression for the energy of the alpha particle is:

$$ E^2 = p^2 c^2 + m^2 c^4 $$

where the relativistic momentum is:

$$ p = \gamma m v $$

And putting $E = 4.5$ pJ we get $v \approx 3.7 \times 10^7$ m/sec or about $0.12c$, which corresponds to an energy of around 19MeV.

The energies of naturally occurring alpha particles such as the ones used by Rutherford are around 5MeV, and this is a factor of four too low for the alpha particle to actually hit the gold nucleus so there were no nuclear reactions in Rutherford's original experiment. However the energy isn't far off, and with a lower charge nucleus the alpha particle energy is great enough to hit the nucleus and react with it. In 1917 Rutherford collided alpha particles with nitrogen and observed transmutation into oxygen with ejection of a proton. Assuming that by strong force you mean the strong nuclear force then this would count as the first strong force interaction with a positively charged particle.

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  • $\begingroup$ One might clarify that your ~5MeV is for naturally occurring alpha particles in the U238 decay series, i.e. those that Rutherford (and others) had readily available at the time. Your calculation is also supported by various handbooks of ion beam analysis (say Tesmer and Nastasi), which has values for when the He4 cross section deviates by 2% (and 4 and 6%) from pure Rutherford, ranging from 1.61MeV for C12 to 7.50MeV for Ge72. Extrapolation to Au gives 18.1MeV, or 19.2MeV for a 6% deviation. $\endgroup$ – Jon Custer Dec 22 '16 at 17:50

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