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Randomness of Radioactivity

  1. Aug 17, 2010 #1
    I understand decay is truly a random process approximated by half-life. So two particles that appear similar to us in every way will experience decay at different times impossible for us to predict.

    Are there any theories that explain some underlying process that dictates this action. Even if it is impossible to know the variables in the formula and thus make a calculation. Is this due to the uncertainty principle?
    Is this it: CKM matrix?
     
  2. jcsd
  3. Aug 17, 2010 #2
    Two assumptions are made both of which are borne out by experiment:
    1.The activity(decays/sec)is proportional to the number of radioactive atoms present.
    As an example two moles of U235 is twice as active as one mol of U235
    2.The activity is characteristic of the isotope
    As an example one would expect that one mol of U235 has a different activity to one mol of
    U238.
     
  4. Aug 17, 2010 #3

    jcsd

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    The most basic kind of model of alpha decay would be to use basic quantum mechanics to find the wavefunction of the alpha particle in a spherically symmetric potential well (the potential models the rest of the atom). The probabilty that the alpha particle has decayed after a certain time is then just the probabilty of finding the alpha particle outside of the potential well at that time. The model is intrinsically quantum as classically you would never find the alpha particle outside of the potential well.

    The model in many ways greatly simplifies the situation, but it still makes predictions that roughly agree with experimental results (more accuarte predictions require more accurate models). Here you can see the randomness of radioactive decay is a result of the randomness of quantum mechanics.
     
  5. Aug 17, 2010 #4
    CKM matrix is not the immediate answer. CKM matrix has to do with the strengths of the weak interaction for quarks in the standard model.

    Because of asymptotic freedom of QCD, it turns out that at low energies the quarks are not free, but actually bound. Nucleons are bound states of three quarks and these are the lowest-energy excitations of the quark field. At low energies, quarks are "invisible" to the experiments. Therefore, it makes sense to define a nucleon field as an effective theory.

    Since protons and neutrons have very similar properties except for the electric charge, Heisenberg had proposed that they are actually two states of the same matter field - let's call it nucleon field. The quantum number associated with the identification of the states of this field is called isospin, in analogy to the spin-1/2 case.

    Then, the problem of nuclear binding is reduced to an effective many-body problem for nucleons interacting through the nuclear force which is independent on isospin and taking Coulomb forces into account.

    Some nuclear states are metastable and they have a finite lifetime [itex]\tau[/itex]. The lifetime and the time constant of the radioactive decay are then related according to:

    [tex]
    \lambda = \frac{1}{\tau}
    [/tex]
     
  6. Aug 18, 2010 #5

    DrDu

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    One should also keep in mind that any Hamiltonian bounded from below cannot give rise to a purely exponential decay law in QM. However, the deviations from exponentiality are much to small for typical decay processes to be detectable.
     
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