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Say you had a very radioactive element in a confined area: could that element (hypothetically speaking) go through beta decay, then, once it has too many protons could it immediately go through electron capture to become that same radioactive element OR become another unstable element that will again go through beta decay?

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Why the confined area? Also, why do you assume that an element that undergoes beta decay would then have too many protons? –  Aesin Aug 24 '13 at 10:19
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become another unstable element that will again go through beta decay?

Yes, this happens. If you start far from the line of stability, many beta decays are needed in order to get to something stable. For heavy elements, you can also have beta and alpha decays intermixed in the chain.

could that element (hypothetically speaking) go through beta decay, then, once it has too many protons could it immediately go through electron capture to become that same radioactive element

There are three processes: $\beta^+$ decay, $\beta^-$ decay, and electron capture. When people refer to beta decay, it means all of these, not just $\beta^-$. Anyway, the kind of process you're referring to is not possible if all the decays we're talking about are beta decays of ground states.

First off, the way you're describing it you seem to be imagining the nucleus oscillating back and forth across the line of stability. Since almost all mass numbers have at least one stable isotope, this can't happen; once you hit stability (in the ground state), you can't decay due to conservation of energy.

Even if you don't cross and recross the line of stability, you can't have a chain like this due to conservation of energy. If the ground state of A can decay to B without violating conservation of energy, then conservation of energy prohibits B from decaying to A.

It might be possible for this to happen if A decayed from a state that was not its ground state, went to B, and then B decayed back to A's ground state. However, this doesn't seem likely to me, and I don't know of any examples. Typically, excited states in nuclei decay electromagnetically on very short time-scales (nanoseconds or picoseconds), so weak decay branches can't compete. To get competition, you typically have to have a state whose electromagnetic decay to the ground state is hindered by the very low excitation energy (gamma decay is slower if the gamma-ray energy is lower), as well as possibly a large difference in spin. This is fairly common in odd-odd-nuclei. However, this only works if you have a low excitation energy in A, which then makes it unlikely that you'd have enough energy left in B to get back to the ground state of A. Furthermore, the staggering of binding energies between even-even and odd-odd nuclei of the same mass number makes it extremely unlikely that the even-even nucleus B would decay back, away from the line of stability, to the ground state of the odd-odd A.

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I don't imagine energy conversion is a problem. Wouldn't it be possible for the initial nuclei to have high enough energy to conduct one or two of these chains, even if the possibility is extremely low? –  namehere Aug 24 '13 at 14:16
    
@namehere: Good point, it might be possible if they're not in their ground states. I'll modify my answer. –  Ben Crowell Aug 24 '13 at 14:21
    
Particular example for multi-beta decay: Pb-210 decays primarily by beta to Bi-210 which decays primarily by beta to Po-210. –  dmckee Aug 24 '13 at 16:02
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