# Why do you need debroglie wavelength less than nucleus diameter to probe a nucleus via deflection? [duplicate]

e.g. you need to give enough energy to electrons to probe a nuclear structure via deflection.

• I googled my own question and got nothing. Or maybe I don't know the specific vocabulary to find the answer. Either way this question comes from reading a book, where a book makes this statement but doesn't explain why(or it expects me to understand why from previous content in the book). Sep 10, 2015 at 19:44
• I dunno, @NeuroFuzzy, this question seem more or less ok (although it's probably effectively a duplicate of a bazillion other questions). Sep 10, 2015 at 21:42
• @DanielSank I had second thoughts and asked about that in chat. You're probably right. Comment deleted!
– user12029
Sep 10, 2015 at 22:07
• I'm not really sure what you're asking
– Sean
Sep 10, 2015 at 23:58
• possible duplicate of Wavelength and resolution Sep 11, 2015 at 13:01

The scattering of waves from much smaller particles is described by Rayleigh scattering, where the intensity of the scattered wave scales as $\frac{1}{\lambda^4}$. If you have one response when the wavelength of light is $100 \mathrm{nm}$, then you will have $\frac{1}{16}$ the response when the wavelength is $200\mathrm{nm}$.
• Alright, so I understand that detecting structures would be more difficult due to the intensity (probability) of scattered particles scaling with $\frac{1}{\lambda^4}$ . The only thing I'm confused about now, is that Reyleigh scattering (from my interpretation of the article) has to do with the EM wave interaction with dipole moments in the particles. I can only see the relation to my question by analogy. Anyways, your answer helps. P.S. does "Rayleigh's criterion for resolvability" come into this? Sep 10, 2015 at 20:31