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I Is nuclear decay probability always constant?

  1. May 4, 2017 #1
    I have heard that the probability of an unstable nucleus decaying is always constant. Is there any way to change this probability?
     
  2. jcsd
  3. May 4, 2017 #2

    mfb

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    In a few exotic cases there is. The two most notable examples:

    - Beryllium-7 can only decay via electron capture. Normally there are electrons around to capture, so it is radioactive. Remove all electrons from its environment and it cannot decay any more - it gets stable.
    - A neutral dysprosium-163 atom is stable. If you remove all electrons, it can beta decay, where the electron stays in a low energy level. The energy is not sufficient to have the electron escape or occupy a higher energy level, which would be required for a decay of a neutral atom.
     
  4. May 18, 2017 #3

    ORF

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    Hello

    There is also another subtle detail: that unstable nucleus, which is going to decay, can be previously excited before decaying; so to be "more" correct, it should be said that the probability of decay is constant for each initial/final state (can be the ground state, or excited states) of the mother/daughter respectively.

    For example, during a supernova, its core can reach enough temperature to excite some isotopes of Sm. The decay probability of these isotopes of Sm is different if they are excited or not. So measuring the decay rate, nuclear astrophysicists can get (approx) the temperature of the core of the supernova.

    In a formal lenguage, the probability of decay is proportional to
    --> a term related to the number of phase states avaibles. So, removing all initial phase states (case of Be-7), or fulling all final phase states (case of Dy-163) you can change the decay probability.
    --> a term related to the nuclear force, about "how strong" the nuclear force can act to change the nucleus (one or more nucleons). If the initial state of the nucleus which is going to decay is different, the decay probability can be also different.

    https://en.wikipedia.org/wiki/Fermi's_golden_rule#The_rate_and_its_derivation

    Regards,
    ORF
     
  5. May 18, 2017 #4

    TeethWhitener

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    To add a few things: as @mfb said, the half-life of Be-7 (and other nuclei that decay via electron capture) is directly (edit:) inversely related to the electron density at the nucleus, so you can increase the half-life by decreasing electron density, or shorten its half-life by increasing electron density--the half-life is sensitive to chemistry.

    I don't know if you'd consider fission by neutron capture a "changing half-life," or if it's better viewed as transmuting a nucleus into a less stable nucleus. But a similar effect also happens with muon capture:

    http://www.phy.bnl.gov/~djaffe/MINOS/References/mu_capture_PhysRep354.pdf

    Fissile elements which capture muons show signs of increased fission due to two processes: a prompt process mediated by excitation of the nucleus by X-rays released as the muon cascades to a 1s atomic energy level, and a delayed process caused by direct capture of muons by nucleons via the weak interaction.

    (Side note: I've long wondered about the naive alpha decay picture of tunneling through the Coulomb barrier and whether the tunneling probability would be affected by the presence of a 1s muon, whose spatially compact wavefunction and negative electric charge may change the profile of the Coulomb barrier in the immediate vicinity of the nucleus. It's probably a minute effect if it even exists at all.)
     
    Last edited: May 18, 2017
  6. May 18, 2017 #5

    ORF

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    Hello

    A decay is spontaneous, and initially there is only the nucleus which is going to decay. If initially there was more than one nucleus, it's called "reaction" instead of "decay".

    @TeethWitener: it's an interesting question. I just found this,
    https://arxiv.org/pdf/1509.09106.pdf

    Regards.
     
  7. May 18, 2017 #6

    mfb

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    Electron capture is called "decay", and the suppression of a decay via adding electrons is changing the decay as well.
     
  8. May 19, 2017 #7

    vanhees71

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    There's an extreme example of astrophysical relevance: A ##^{187}\text{Re}## atom has a half-time ##T_{1/2}=42\cdot 10^9 \mathrm{yr}##, while for the totally ionized nucleus it can decay into a bound atomic state (which is Pauli blocked for the atom), which leads to a half-life of ##T_{1/2}=14 \;\mathrm{yr}##. This is about 10 orders of magnitude smaller! It's of astrophysical relevance, because the Re/Os clock is used to determine astronomical ages like the age of our solar system, but it's not so safe considering that the Re is ionized in stars, so that the half-life is not in the billion-year range but lower by more than 9 orders of magnitude!

    http://iopscience.iop.org/article/10.1238/Physica.Topical.080a00028/meta
     
  9. May 19, 2017 #8

    ORF

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    fHello

    Very interesting, thank you! :)

    @mfb : mmm... tousché! I think about another counter-example: fission by Coulomb breakup (no nuclear-interaction) is a nuclear reaction or a decay?

    As far as I know (I can be wrong), the point is that the nucleus can reach several nuclear states. These nuclear states are connected by the nuclear interaction + external fields (pressure of nuclear matter, electric field, etc) + chemical potential (beta-stability in neutron stars). The same idea as in thermodynamics.

    So, if you don't have external fields (or are negligible), and the nucleus is not affected by the chemical potential of electrons or other external particles, can the change of the nuclear statebe called "spontaneous decay"?

    Example: beta-deay is a readjustement of the chemical potential of the nucleons, and this relationship should be accomplish

    $$\mu_n = \mu_p + \mu_{e^-} + \mu_{\nu e}$$ (in extreme cases, you have to take into account also muons)

    So, if the chemical potential of neutrino or electron are not negligible (ie, they can not go away/in freely), they will tend to modify the nucleon equilibrium through the chemical potential term.

    I don't know if there is a clear difference between decay an reaction.

    Regards,
    ORF
     
  10. May 19, 2017 #9

    mfb

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    How else would you call it?

    I don't think it helps to spend too much time on defining words. Just use something that is clear.
     
  11. May 20, 2017 #10

    TeethWhitener

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    Thanks for finding this. It's pretty much exactly what I had in mind, up to and including applications to nuclear waste mitigation, except the author considers (far more relevant and feasible) electrons rather than muonic atoms.
     
  12. May 22, 2017 at 10:03 AM #11

    ORF

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    Hello

    So, to sum up and give a final answer (please, confirm if you agree, or not)

    Question: Is nuclear decay probability always constant?
    Answer: the probability of decay will be given by the Fermi Golden Rule (approx). The probability can be affected via phase space, via Hamiltonian of the interaction (nuclear Hamiltonian + external fields term + chemical potential term). So, the probability depends on the Hamiltonian, the initial and final state of the nucleus and the density of states in the phase space. If this items are the same, the probability will be the same. If something changes, the probability changes.

    Please, can someone confirm this?

    Thank you for your time (and thank you for your question, it's a good one :) )

    Regards,
    ORF
     
  13. May 22, 2017 at 10:22 AM #12

    mfb

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    Sure.
    That sounds quite trivial.
     
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