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Binding energy calculation of a nucleus

  1. Jan 19, 2010 #1
    Hello,

    I'd appreciate feedback on my calculation of binding energy, just wanted to check if I'm on the right track w/ regards to calculation steps (and thought process).

    I'll try to be as clear as possible, please give me input if there's some notations / steps I should correct and keep in mind I have limited knowledge on the subject.


    1. The problem statement, all variables and given/known data
    Basically I want to calculate the binding energy per nucleus of the isotope 11B in MeV.

    Binding energy of the nucleus, which is the energy keeping the protons and neutrons together, thus the energy needed to break apart the "bonds" between the p+'s and n0's.
    (in terms of force notation, would this be the strong nuclear force?)


    Data given:

    11B: m = 11,0093 amu (Z = 5) (Z = # of p+ (I assume, since the # of p+ relates to the "iso" part of isotope))


    mass/proton = 1,007276 amu
    mass/neutron = 1,008665 amu


    2. Relevant equations

    ∆m = (sum of protons and neutrons) - (measured mass of the isotope)

    E=mC2

    binding energy = ∆m·C2

    3. The attempt at a solution

    11B ∑m = 5p+ * 1,007275 amu + 6n0 * 1,008665 amu = 11,08837 amu (the unbound system calculated mass)

    11B m = 11,0093 amu (measured somehow)


    ∆m = 11,08837 - 11,0093 = 0,07907 amu

    Binding energy = 0,07907 amu * (3E8 m/s)^2 = 7,1163E15 J ( is it safe to say this is joules kg·m^2/s^2 ?) or do I need to convert amu's into Kg's first?

    Assuming I get the result in J I continue to convert into MeV:

    1 J = 6,24E18 eV = 6,24E12 MeV

    MeV: 7,1163E15*J * 6,24E12 MeV/J = 4,44E28 MeV

    This seems to me like a LOT of energy, how can I best evaluate my answer? Does this amount of energy seem like a reasonable amount? Can I relate it to the strong nuclear force?

    Thanks.
     
  2. jcsd
  3. Jan 19, 2010 #2
    I looked up some binding energy values, and they all lay in quite a different interval. That value I got has to be horrendously off course.

    Maybe I should divide by the atom number, but then again I would still get a pretty crazy value.
     
  4. Jan 22, 2010 #3
    I think you're more likely to get a response to this in the physics sections for homework questions (don't ask me where!). I don't think you can post the same question twice though, you may like to ask a moderator to transfer it, perhaps.
     
  5. Jan 23, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    Yes.

    --
     
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