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Alpha decay potential barrier.

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A nuclei of a atomic number Z decays into a alpha particle (a He nucleus with Z =2) and a daughter nucleus with [itex](Z_{d}[/itex]).
    The decay may be described as the tunneling of an alpha-particle through a barrier caused by the Coulomb potential between the daughter and the alpha-particle.

    The potential is: [itex] V(r) = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{r} [/itex]



    Knowing that the probability of transmission , T, is proportional to

    [itex] T \propto e^{{-2* \int_a^b \sqrt{\frac{2m(V(r)-E)}{\hbar }} \,dr }} [/itex]

    Calculate the turning points, [itex]a[/itex] and [itex]b[/itex].



    Notes: The diagram for the potential barrier is shown inf the link:
    https://www.google.pt/search?q=alph...taneous_Decay_Processes_-_Alpha_Decay;624;354


    Im stuck in this problem. I know that i can calculate the turning points at V(r) = E(r), but i do not have E(r). Can u give me a tip for solve this problem?
     
  2. jcsd
  3. Mar 31, 2013 #2
    I found the second turning point doing the following:

    E(r_{2})=V(r_{2})

    so


    [itex] V(r_{2}) = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{r_{2}} [/itex]

    then we [itex]r_{2} = \frac{1}{4\piε_{0}}\frac{2Z_{d}e^{2}}{E(r_{2})}[/itex]
     
    Last edited: Mar 31, 2013
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