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Semi empirical mass formula

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data
    Using semi empirical mass formula calculate which isotope of oxygen (##_8O##) decays into mirror nuclei. Is it ##\beta +## or ##\beta -##?

    2. Relevant equations
    $$W(A,Z)=w_0A-w_1A^{2/3}-w_2\frac{Z^2}{A^{1/3}}-w_3\frac{(A-2Z)^2}{A}-w_4A^{-3/4}\delta(A,Z)$$ where $$w_0=15.8MeV$$ $$w_1=17.8MeV$$ $$w_2=0.71MeV$$ $$w_3=23.7MeV$$ and $$w_4=11.2MeV$$

    3. The attempt at a solution

    Hmmm, firstly: $$\beta ^+ : \qquad ^A_8O \rightarrow ^A_7X+e^++\nu _e$$ and $$\beta ^- : \qquad ^A_8O \rightarrow ^A_9Y+e^-+\bar{\nu _e}$$ I don't really know what to do here but..... My idea was that maybe the energy balance would give me the answer?...

    Something like:
    For ##\beta ^+##: $$M_Oc^2=M_xc^2+m_ec^2$$
    $$8m_pc^2+(A-8)m_nc^2-W(A,8)-(7m_pc^2+(A-7)m_nc^2-W(A,7)+m_ec^2)=C$$ Now my idea was if that ##C>0## than the decay is not possible. But... I amnot sure this is the right way, since I don't have the number ##A##.

    Does anybody have a hint on what to do here? :/
     
  2. jcsd
  3. Apr 7, 2015 #2
    You want to show (as you stated) that the mass of the left side of the decay equation is greater than the mass on the right side. You must choose the value of A for known isotopes of oxygen and try each one to see what happens.
     
  4. Apr 7, 2015 #3
    Aha, so there isn't any mysterious way to find that A. I just have to guess it and look for the right one?
     
  5. Apr 7, 2015 #4
    Yep. Common isotopes of oxygen are 14,15,16,17,18,19, 20 .three are stable. There are a few more with Half-Lifes less than 1 sec
     
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