How Electrons Penetrate Nucleus w/o Capture

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http://en.wikipedia.org/wiki/Radioac...ng_decay_rates

"...which are subject to electron capture in 7Be because (like all s atomic orbitals in all atoms) they naturally penetrate into the nucleus."

How can the s orbitals penetrate the nucleus without being captured? And do they mean all energy levels of the s orbitals?
 
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S-wave electrons do penetrate the nucleus, but may or may not be subject to capture. If and when an orbital electron is captured it turns a proton into a neutron and emits a neutrino. Whether this can actually occur depends on the energy available - does the resulting nucleus have more or less binding energy than the original one?
 
If I'm not mistaken, that would depend on if the resulting nucleus is heavier than Nickel-56, (or 57? something like that). Lighter and I believe it would have less binding energy and heavier the opposite. Or does the fact that there are still the same number of nucleons mean that it stays the same?
When I think of a nucleus I think of a solid chunk which can't be penetrated which is why I asked the question but I can see that's not the case. When thinking about the capture event, can we think of a proton as one unit or does it have to hit a specific quark? And while I'm on the subject of a nucleus, what are the distance between nucleons? Are nucleons really triangles like they are depicted when showing quarks?

http://upload.wikimedia.org/wikipedia/commons/c/cf/HAtomOrbitals.png
This picture suggests to me that the electrons would have to almost teleport in some situations. The 3s,3p, and 3d orbitals seem to have much more defined areas where an electron can't go. Is there a somewhat simple explanation for this? Or does this get into complicated quantum effects only present on that scale?
Sorry for all the questions, I actually find this stuff really interesting!
 
You might find this tutorial helpful. It covers electron capture as well as both kinds of beta decay.
 
samblohm said:
... When I think of a nucleus I think of a solid chunk which can't be penetrated which is why I asked the question but I can see that's not the case...
A better simplistic visualisation is a cloud. The quarks and gluons within each nucleon are all whizzing around, because the uncertainty principle requires their momenta to be significant because of the very small uncertainties in their positions.
...When thinking about the capture event, can we think of a proton as one unit or does it have to hit a specific quark?
It does, indeed, interact with a single, up, quark. At this level, the reaction is

e + u → v + d​

Putting in the W boson, the full process is actually two interactions:

e → v + W-
W- + u → d​
...And while I'm on the subject of a nucleus, what are the distance between nucleons? Are nucleons really triangles like they are depicted when showing quarks?
≈ 1 fm. They are not triangles, see above re clouds.
 
Is there any order that the quarks and gluons move around in nucleons?
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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