A question about a particle's de Broglie wavelength

[weezy]

For electron scattering experiment which measures charge radius of the nucleus, it's said that energies of scattering electrons (or protons) must be such that their wavelength size is of the same order as nuclear dimensions. While i understand why this must be true I'm not entirely sure. 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. 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 not sure if this is the case hence I'm seeking for clarification. Thanks.

 

[Paul Colby]

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

This is common to all wave like phenomena. Laser light and radio waves are the same phenomena, electromagnetic waves. The best resolution of a microscope is determined by the wavelength of light (and other practical considerations). Radar's resolution is limited to some fraction of the wavelength which is measured in feet.