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Binding enregy per nucleon

  1. Aug 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the binding energy per nucleon of
    238
    92 U in MeV

    2. Relevant equations
    ΔM= ideal mass - atomic mass
    Ebind= ΔM * 931.5



    3. The attempt at a solution

    first i found the diference in mass from ideal weight and atomich weight which is 1.98499. then i multiplied it by 931.5 which gives me a total binding energy of 1849.0181 MeV.
    Now, i understand that a nulceon is a pair made of a proton and a neutron so i should divide my answer by the number of nucleons in the atom. but there are more neutrons than protons in U so i dont know what should i divide by!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 19, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    "In chemistry and physics, a nucleon is one of the particles that makes up the atomic nucleus. Each atomic nucleus consists of one or more nucleons, and each atom in turn consists of a cluster of nucleons surrounded by one or more electrons. There are two kinds of nucleon: the neutron and the proton. The mass number of a given atomic isotope is identical to its number of nucleons. Thus the term nucleon number may be used in place of the more common terms mass number or atomic mass number." http://en.wikipedia.org/wiki/Nucleon

    ehild
     
  4. Aug 19, 2012 #3
    oh, it makes sence now, thanks!
     
  5. Aug 19, 2012 #4

    CAF123

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    Gold Member

    Total binding energy, [itex] E_b = (Zm_p + Nm_n - M(Z,N))c^2 [/itex]
    You should find that, in adding up all the constituent nucleons that their sum is greater than that of the mass of the actual nucleus. This 'extra' mass is the binding energy.

    The binding energy per nucleon [itex] = \frac{E_b}{A}, [/itex] where A is the number of nucleons (number of protons and neutrons)

    EDIT: I see that your question has already been answered
     
  6. Aug 19, 2012 #5
    yea haha, thanks anyway
     
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