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Q-value calculation - nuclear physics

  1. Feb 26, 2008 #1
    http://www.nndc.bnl.gov/masses/mass.mas03, and the missing data:
    u=931.50MeV
    p=1.00728u=938.28MeV
    d=2.01355u=1875.63MeV
    e=0.511MeV

    A(a,b)B=a+A->b+B
    Q=[(m(a)+m(A))-(m(b)+m(B)]MeV (1)

    This can't be more simple, but in every example that I saw in my lectures different values occur. For example:
    3He+3He->4He+2p (this is not 3 times He, but p+p+n), Q=12.859
    Using the table of masses from a given link and (1) I get Q=13.83MeV.

    The second thing I don't understand is when I have this situation:
    d+p->3He+photon,
    1875.63MeV+938.28MeV->2809.4MeV+photon. Is Q now equal to energy of photon?
    In this case 1875.63 + 938.28 - 2809.4=4.51MeV+photon.
    Btw, also in my lectures Q of this reaction is 5.493MeV.

    I really can't understand these differences.
     
  2. jcsd
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