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Are the other peaks with lower energy caused by the possibility that daughter nuclei have to be in excited states?as show in this link (count versus energy)

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    $\begingroup$ Usual 'daughter' nuclei in American English usage. But, yes, there are discrete transitions from one nuclear energy level to a different nuclear energy level (unless it is a 3-body decay which messes it all up). $\endgroup$ – Jon Custer Jun 15 '15 at 19:24
  • $\begingroup$ it's discrete due to the fact that is an two-body decay isn't it? $\endgroup$ – physnolimits Jun 15 '15 at 19:28
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    $\begingroup$ You are going from one energy level in one nucleus, to one (or several) energy levels in another. Then, being only a 2-body problem means that energy and momentum conservation is one unique alpha energy. Going 3 body smears things out... $\endgroup$ – Jon Custer Jun 15 '15 at 19:50
  • $\begingroup$ @JonCuster Please convert your comment to an answer. $\endgroup$ – rob Jun 15 '15 at 23:50
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Nuclei have a series of discrete energy levels (somewhat analogous to electronic energy levels, but the details are, not surprisingly, different). Examples of these so-called Energy Level Diagrams can be found at, e.g., Triangle Universities Nuclear Laboratory. So, a simple alpha decay will go from one level in the parent nucleus to one level in the daughter nucleus, releasing a set amount of energy. If it is pure alpha decay, than it is a 2-body problem so there is a unique solution to conserve energy (released) and momentum (net zero), so the alpha comes out at a single energy. (If there are multiple final energy states of the daughter nucleus, than there may be several possible alpha energies).

Contrast this to beta decay, where the broad energy spectrum of the resulting electron was one of the clues to the existence of the neutrino. Here, the end state is a three body problem so the released energy and momentum can be parceled out amongst the products in many ways.

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