# 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). – Michael Sep 10 '15 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). – DanielSank Sep 10 '15 at 21:42
• @DanielSank I had second thoughts and asked about that in chat. You're probably right. Comment deleted! – user12029 Sep 10 '15 at 22:07
• I'm not really sure what you're asking – Sean Sep 10 '15 at 23:58
• possible duplicate of Wavelength and resolution – ACuriousMind Sep 11 '15 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? – Michael Sep 10 '15 at 20:31