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Why alpha and beta?

  1. Dec 19, 2009 #1
    Do we know why radioactive nuclei only decay through emission of alpha and beta particles? There is spontaneous fission in some cases, of course, but what I mean to ask is, why should just a proton or just a neutron or a particle X with 2 protons and 3 neutrons, or whatever else not be emitted? Why does this particular combination of 2 and 2 in alpha particle work as against any other?
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  3. Dec 19, 2009 #2


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    cluster decay

    Hi ksac! :smile:

    From http://en.wikipedia.org/wiki/Cluster_decay" [Broken] …
    … see the full article for more details. :wink:
    Last edited by a moderator: May 4, 2017
  4. Dec 19, 2009 #3


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    The alpha particle is a pretty stable collection of nucleons.

    Radionuclei also experience electron capture in which a K-electron (and possibly L-electron) will be capture by the nucleus and a proton will be transformed into a neutron. It has the same effect as positron emission, in which Z decreases by 1 with A (integer) remaining the same.

    Also, some of the transuranics undergo spontaneous fission.
  5. Dec 19, 2009 #4


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    Beta decay is qualitatively different from alpha decay. It's a weak-force process, and it doesn't involve emission of a particle or cluster of particles that were already present inside the parent nucleus.

    Tiny-tim has pointed out that other types of particle emission do occur. Another thing to realize is that in a hot nucleus (e.g., a nucleus that's just been created via fusion reactions in an accelerator), you do get lots of proton and neutron emission.

    The reason you'll never see spontaneous neutron emission in a cold nucleus is that there's no Coulomb barrier to tunnel through. Therefore any neutron that has enough energy to escape will do so immediately, not after a longer time delay as with charged particles.

    The main reason that alpha decay tends to be favored is that the alpha is a doubly magic nucleus, so it has an extremely large binding energy. That means that by phase space arguments, alpha emission is strongly favored.

    Another issue is the somewhat foggy notion of "preformation." A straightforward way of estimating alpha decay rates is as follows. You calculate the probability P of tunneling out through the coulomb barrier. You estimate the frequency f with which an alpha particle assaults the barrier, which is basically found by taking the velocity scale (1% of the speed of light) and dividing by the distance scale (a few fm). Then 1/(Pf) gives an estimate of the alpha decay half-life. This estimate has all the right behavior as a function of A and Z, but it's much too short by some large, constant factor. To account for this you assume that there's only some small probability that neutrons and protons will already have gotten together and formed themselves into an alpha particle. This preformation probability is difficult to calculate, but it makes sense that it's higher for alphas than for other clusters. This is because (a) an alpha is small (only 4 particles), (b) the nuclear residual interaction includes a strong pairing force, which sort of favors the formation of nn and pp correlations, and (c) the binding energy of an alpha is very favorable. This is really just a plausibility argument. Formally, you really can't have an alpha particle cruising around inside a nucleus, because of the Pauli exclusion principle. However, nucleon-nucleon correlations inside nuclei do exist.
  6. Dec 19, 2009 #5
    Sometimes a radioactive nucleus will emit a neutron along with an antineutrino. This the process called K-capture (where an orbital electron and a proton convert to a neutron) discussed by Astronuc above.
    Bob S
  7. Dec 19, 2009 #6


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    Careful - In electron capture, the neutron stays in the nucleus. Neutron emission may happen when a gamma ray of sufficient energy interacts with the nucleus - a process known a photo-neutron emission - or photodisintegration in the case of a deuteron (d (γ,n) p).

    One can also look at the requirement for various decay schemes - looking at the masses of the nuclei before and after the reaction with E or ΔE = Δm c2 to see if E or ΔE is > 0 (exothermic, preferred) or < 0 (endothermic, not spontaneous, but requires energy input).
  8. Dec 19, 2009 #7
    My bad Sorry Bob S
  9. Dec 21, 2009 #8
    I forgot to mention two neutron-emitting radioactive sources I used as a grad student: Plubyllium (symbol PuBe), and Pobyllium (symbol PoBe). I don't know what the plutonium polonium beryllium mixture ratios were. Lots of neutrons, but the alphas were fully contained.
    Bob S
    [added] Read the abstract of
    Last edited: Dec 21, 2009
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