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Binding energy of Li 6

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    How do I find the binding energy for the nuclei of Lithium 6.
    N=Z=3, atomic mass=6.015 u, binding energy per nucleon= 5.33 MeV
    The equation I'm using is :

    B(N,Z)=Z*m(0,1)*c^2+N*(mass of n)*c^2-m(N,Z)*c^2.

    B(N,Z)=3*(1.00783u*1.66E-27kg)*(3E8 m/s)^2+3*(1.008664u*1.66E-27kg)*(3E8 m/s)^2-(6.015u**1.66E-27kg)*(3E8 m/s)^2

    B(N,Z)= 4.84056E-12 J *6.24E18 eV= 30.2 MeV

    but that's not the answer, I also tried dividing that answer by the binding energy per nucleon which is 5.332 MeV and I got 5.66 MeV that's not the answer. I'll appreciate it if someone explained to me what I'm doing wrong. Thanks.
     
  2. jcsd
  3. Nov 26, 2011 #2
    You are given the binding energy per nucleon and you know the number of nucleons so you should be able to get the binding energy of the NUCLEUS.
    Is that what the question means?
     
  4. Nov 26, 2011 #3
    Well, you are right, I didn't catch that , now I got 31.98 MeV, which is close to the same answer I got doing all the needless calculations but not quite, since now I got it right. Thanks!
     
  5. Nov 26, 2011 #4
    that is good to hear !! sometimes we are blinded by information!!
     
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