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Homework Help: Alpha Decay

  1. May 30, 2012 #1
    I am completing an assignment that is covering alpha, beta, and gamma decay. I am going to try and keep this as general as possible, as I want to figure this out myself but I just looking for feedback to make sure I am on the right track.

    I noticed that after alpha decay, the mass of the alpha particle and daughter atom is not quite equal to the parent atom. It wants to know whether I think mass and energy are conserved relative to E=mc^2. I noticed there was roughly 8.0x10^-13 J's less after the reaction. But I did a bit of research on the speed of an alpha particle, converted the mass of the alpha particle to kg, and when I plugged them into the kinetic energy formula (ek=.5mv^2), it gave roughly the difference between the parent vs. daughter alpha particle. That being said, could some of the mass have been converted into kinetic energy of the alpha particle?
  2. jcsd
  3. May 30, 2012 #2


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    Yes! I believe you're right. :approve: Very nice.

    The energy release of alpha decay (difference between the rest mass of the initial atom minus the sum of the rest masses of the daughter atom and alpha particle, times c2) is primarily in the form of kinetic energy.

    And most of that kinetic energy is in the motion of the alpha particle, particularly if the parent atom was heavy. Due to conservation of momentum, the daughter atom might end up with a small fraction of that kinetic energy, but most of the energy is with the alpha particle. (You can prove this to yourself as an exercise. Using conservation of momentum, find the respective velocities of two objects separating from each other, initially at rest, when one mass is much bigger than the other. Then find the kinetic energy of each object via ½mv2*, and compare. You should find that the less massive object gets more of the energy.)

    *(Using the Newtonian K.E. = ½mv2 should approximate this pretty well. The velocity of alpha particles is a pretty small fraction of c.)
  4. May 31, 2012 #3

    Is it also safe to say that that velocity of the alpha particle will be different in some isotopes due to the different sizes and charges? (referring to Coulombe's Law)
  5. May 31, 2012 #4


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    I'm not an expert on this subject, but I'm led to believe that there is surprisingly little variation in the decay energy. (And for the most part, a parent atom has to be pretty heavy in the first place [heavier than maybe nickel?], before it is likely to emit an alpha particle. Meaning that most kinetic energy stays with the alpha particle rather than the daughter atom, for even lighter parent atoms [relatively speaking]. So there's not a whole lot of variation.) I've provide a link below regarding the "Geiger–Nuttall law" that might be informative.

    Last edited: May 31, 2012
  6. May 31, 2012 #5
    Since we're on the subject of decay,

    In beta plus decay, and electron is released, but within the daughter atom a proton will fuse with an electron to create a neutron. That being said, the mass of the daughter atom is a bit larger. I'm guessing that what when the proton fuses with the electron, the energy that kept the electron in orbit around the atom was converted to mass along as well? What are your thoughts?
  7. May 31, 2012 #6


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    I think you mean a positron (and a neutrino).
    don't confuse β+ decay with electron capture decay. β+, β-, and electron capture decays all have similarities, but they are not the same thing.
    Try that one again.

    Let me give you an example of a β+ decay:

    40K40Ar + e+ + νe

    (where νe is a neutrino).

    Now compare the masses of the original Potassium atom to that of the Argon atom:

    Potassium 40 mass (in unified atomic mass units): 39.963998475 u
    Argon 40 mass (in unified atomic mass units): 39.96238312251 u​

    (Source WolframAlpha, http://www.wolframalpha.com/)

    The difference in that mass energy (ΔE = Δmc2) becomes the mass energies of the positron and neutrino plus their kinetic energies (plus a little involving the recoil of the Argon atom).


    And unlike alpha decays, the beta decay energy can vary greatly. Not only that, beta particles normally travel at ultrarelativistic speeds, so you can't use the KE = ½mv2. Instead, you need to use the relativistic version, [itex] KE = mc^2(\gamma - 1) [/itex] where [itex] \gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}} [/itex]
    Last edited: May 31, 2012
  8. May 31, 2012 #7
    Interesting, thanks for the info.
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