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Beta decay

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  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    3H has more tightly bound than 3He. Why is it, then, that 3H beta-decays to 3He?

    2. Relevant equations
    I also wonder what is the physical meaning of the value mass excess Δ(Z,A)≡[M(Z,A)-A]c2, or it is just for the convenience of calculations.

    3. The attempt at a solution
    B/A of 3H is 2.8273 MeV while the B/A of 3He is 2.5727 MeV.
    We know that the requirement of beta decay is m(Z,A)>m(Z±1,A)+me, and m is the mass of nucleus because beta decay is the process of nucleus transition, not the process of atom transition.
    I guess the reason that 3H beta decays to 3He is just because the m(1,3)>m(2,3)+me, regardless of whether the daughter nucleus is more bound than mother nucleus. So beta decay has no direct relationship with the binding energy, which means there is also no directly relationship between beta stable line and binding energy, is it?

    Thank you for you time!
     
  2. jcsd
  3. Mar 22, 2016 #2

    TSny

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    Hello and welcome to PF!

    I think your comments are good. The magnitude of the mass excess, expressed in energy units, is the energy required to break the nucleus into individual separated nucleons. Actually, the way you defined the mass excess makes the mass excess negative. So, it's often called the mass defect. Breaking a nucleus apart is different than a nuclear decay process. So, as you say, knowing the mass excess of parent and daughter does not, by itself, tell you if a decay is energetically possible. The energy released in the decay is also important.

    It might help to consider the following procedure. First, break apart the 3H nucleus into 1 proton and 2 neutrons. Think about the energy you would need to supply to do this. Then let one of the neutrons decay into a proton and electron (and anti-neutrino). How much energy is released in this step? Finally, imagine forcing the 2 protons and 1 neutron together to form 3He. How much energy is released in this third step? How does the total energy released in steps 2 and 3 compare to the energy input in the first step? Is the overall process energetically favorable?
     
  4. Mar 23, 2016 #3
    Hi, TSny!
    Thank you very much for your reply!
    You mentioned that the magnitude of mass excess is the energy required to break the nucleus into individual separated nucleons, and it's often negative. But I also find there are some nuclei whose mass excess is positive, such as M.E. of 3He is 14.931 MeV and it is stable, while that of most "heavy" nuclei is negative. Are the light nuclei exceptional?

    I followed your procedure.
    The energy needs to be more than the binding energy 8.4819 MeV.

    E=mn-mp-me=0.782 MeV

    The energy released per nucleon is also the binding energy 7.7181 MeV

    The energy released in the 2 and 3 step is 8.5001 MeV, which is a little more than the energy 8.4819MeV needed to break apart 3H. If I think right, it is energy favorable. So, the beta decay will happen as long as the process is energy favorable whatever the binding energy (mass excess) is.
     
  5. Mar 24, 2016 #4

    TSny

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    I must apologize for thinking that mass excess is the same as mass defect except for sign. As you can see, I now have egg on my face o:).

    Actually, sometimes people do think of them as the same: see https://en.wikipedia.org/wiki/Nuclear_binding_energy#Mass_defect
    But you clearly defined mass excess by the formula you gave and I should have noticed that it was different than mass defect.

    Mass defect is the same as binding energy when expressed in the same units. However, mass excess Δ as defined by your formula is something different. Apparently, some tables of nuclides prefer to specify the mass excess rather than the atomic mass itself. Using your formula, it is easy to switch between atomic mass and mass excess using the value of A for the nuclide.

    Because the atomic mass of carbon-12 is defined to be exactly 12 atomic mass units, you can check that the mass excess of C-12 is zero: Δ = 0. For other nuclides, Δ can be positive or negative. A negative value generally indicates that the nuclide has a greater binding energy per nucleon than C-12. A positive value of Δ indicates less binding energy per nucleon than C-12.

    When you have a nuclear reaction which transforms one or more nuclei into one or more other nuclei, the total A value will be conserved. So, the difference between the initial total Δ and the final total Δ will be the same as the difference in initial and final total masses. This means that when you want to calculate the energy released in a reaction, you can work with mass excess in place of atomic mass. Examples here: https://people.nscl.msu.edu/~schatz/PHY983_13/Lectures/mass_notes.pdf

    Yes, it's energy favorable. But just because it's favorable from an energy point of view doesn't mean that it will actually happen. I think that there could be other factors involved.
     
  6. Mar 27, 2016 #5
    TSny, hello!
    You are so funny. I guess you must enjoy studying physics very much.
    I see your view, and I totally agree with your ideas.
    Thank you very much for your help:smile:
     
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