For the bulk of people, no question : an alpha nucleus is just few femtometers large. That's all.
However, taking the emission of 232Th, it has in most cases a kinetic energy of 3.994 MeV.
Its speed is 4% of light, so we accept the non-relativistic approximation.
Group-speed : 13 877 060 m/s,
Momentum : 9,2210 . 10-20 kg.m/s
so de-Broglie-wavelength : 7,1821 fm/cycle.
The experimental uncertainty on the energy is 1/4000, so uncertainty on the momentum is 1/8000. So I expect that the undefinition is so, or less.
So this wave must be approximately 8000 wavelengths long (or more) , that is 42 pm, 0.42 Å, with a duration of 4 attoseconds, seen by the emitter.
At a de-Broglie-frequency of 9,012886 . 1023 Hz (rest-frequency, non-relativistic approximation).
And it takes about 300,000 ionizations to entirely stop this alpha ray.
These results are very different from the standard believings.
But are there some experimental results that could solve the dilemma ?
And I have no precise ideas on the width of an alpha ray.
May be better to consider spherical rather than square or rectangular having length and width. Use R=R0A^1/3 as sphere. Nuclear forces are spherical symmetric as follows from liquid drop model.
Consider nuclear as round rather than of other shape. So Compton wavelength will be mostly know circular former.
However the experiment has been done of Young holes, with ultra-cold helium atoms. So each one passes through the both holes. Nuclear forces do not tell us all what is to know. And an alpha nucleus has a charge, so it has electromagnetic interactions during all its trajectory.
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