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Kinetic energy of electron after beta decay

  1. Aug 6, 2012 #1
    Is that a constant? If not, is there an equation that allows us to calculate its kinetic energy based on the properties of the atom it came from?
     
  2. jcsd
  3. Aug 6, 2012 #2

    jtbell

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    The total energy available in the decay is fixed, according to the difference in mass between the initial and final nuclei. However, that energy is distributed between two outgoing particles, the electron and the (anti)neutrino, and that distribution varies randomly from one decay to another. So beta-decay electrons from a particular isotope have different energies, ranging from zero up to the total energy available.

    http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/beta2.html
     
  4. Aug 6, 2012 #3

    QuantumPion

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    It is not a constant, it depends on the nuclide. Tritium decays with a 5 keV beta. C-14 decays with a 50 keV beta. Co-60 decays with a 300 keV beta, etc. A description of the beta energy spectrum can be found here: http://hyperphysics.phy-astr.gsu.edu/Hbase/nuclear/beta2.html
     
  5. Aug 6, 2012 #4
    Ok, I think I see. So is this why a tritium decay into He4 isn't very energetic but a decay in a heavier element could be quite energetic?
     
  6. Aug 6, 2012 #5

    QuantumPion

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    heh you beat me to it :)
     
  7. Aug 6, 2012 #6
    Sorry, was typing that question as you were typing the answer :)
     
  8. Aug 6, 2012 #7
    the energy in beta decay is shared between the electron (positron) and a neutrino or antineutrino so jtbell answer is best
     
  9. Aug 6, 2012 #8
    Another newb question, the total energy resulting from a tritium atom undergoing beta decay is 18.59 keV. According to wiki the average kinetic energy of the electron is 5.7 keV leaving 12.89 keV on average attributed to the antineutrino.

    Are those the only particles emitted during beta decay?

    Maybe I did the maths incorrectly but the 12.89 keV kinetic energy translated to relativistic velocity for a particle with the rest mass range of an electron neutrino (< 2.2 eV) puts the upper velocity limit at > 2.998*10^8 m/s. How is that possible?
     
  10. Aug 6, 2012 #9

    jtbell

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    We can't tell you whether your equations are correct if you don't tell us what they are. :smile:

    Even if your equations are correct, you have to be very very careful about the arithmetic in a calculation like this, because roundoff errors in intermediate steps can bite you hard.

    For K = 12890 eV and mc2 = 2.2 eV I get v/c = 0.999999985.
     
  11. Aug 6, 2012 #10
    Small correction, T decays into He3, not He4.
     
  12. Aug 6, 2012 #11
    That's the problem, I don't know the equations, I'm a total amateur :) I was plugging the values into wolfram and that's what it spit out. I think the math it did was bugged because I put in a value of 1 MeV kinetic energy and it was less than 1c.
     
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