For electron scattering experiment which measures charge radius of the nucleus, it's said that energies of scattering electrons (or any other particle) must be such that their wavelength size is of the same order as nuclear dimensions. 

I'm not entirely sure how this works. My understanding is that when the de Broglie wavelength is larger than the nucleus the wave packet is more spread out in space and thus chances of striking and getting scattered by the nucleus is greatly reduced although non-zero.

This should happen because with a larger wavepacket the probability of finding the electron right where the nucleus is located when the wavepacket passes through or engulfs the nucleus decreases with increase in the wave packet size and since probability current density is conserved, the chances of hitting the nucleus diminishes.

With a smaller comparable sized wavepacket we can say almost with 100% guarantee that the electron must be somewhere in the wave packet which totally fits the nucleus this time and hence the scattering occurs.

I'm seeking clarification, as
I'm not sure if this is the case.

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    $\begingroup$ @countto10 i approved your edit. Thanks for your help! $\endgroup$ – Weezy Jun 30 '17 at 6:55
  • $\begingroup$ I have two points that hopefully someone can correct if they are wrong. When you say "striking" , it implies to me that you are still, sorta, kinda, unconsciously...thinking classically (sorry :) But a scattering event, as we both know, is mediated by photons, rather than direct proximity of the particles. The particle as wavepacket idea also has a major flaw in the fact that it disperses extremely quickly, en.m.wikipedia.org/wiki/Wave_packet. I do accept you are referring to the probability wave here as regards wavepackets, but I thought I should mention it in passing. $\endgroup$ – user154420 Jun 30 '17 at 7:41

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