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Nuclear Binding Energy

  1. Jul 12, 2012 #1
    Okay I'm taking an Individual Learning Course in Ontario for SPH4U and this is a Support question:

    How much energy would be required to remove one neutron from nitrogen-14 isotope, given these masses?
    a) N-14 isotope 14.0031 u
    b) N-13 isotope 13.0057 u

    2. Relevant equations
    Knowns
    Electron = 0.000549 u
    Proton = 1.007276 u
    Neutron = 1.008665 u

    3. The attempt at a solution

    Okay so the first step is to determine how many Protons and Neutrons there are: Periodic table says 7. Therefore A = 14, 13 and Z = 7
    Second Step Determine what the unified mass of N-14 without its Electrons
    mnu1 = mN-14 - 7me
    mnu1 = 14.0031u - 7(0.000549)
    mnu1 = 13.999257 u
    Third Step Determine what the unified mass of N-13 by adding its Nucleus components together
    mnu2 = 7mp + 7mn
    mnu2 = 7(1.007276) + 7(1.008665)
    mnu2 = 14.11587 u
    Now we need to subtract the two to find the difference in mass
    |mnu1 - mnu2| = |13.999257 u - 14.11587 u| = 0.102607 u
    We can now convert this into the total binding energy
    0.102607 u (931.5 MeV/c2/u) = 95.58 MeV/c2
    If we divide this by A we will get our average Binding Energy
    6.83 MeV/c2 per Nucleon

    When I do the same procedure for N-13 I get an answer of 7.24 Mev/c2

    The back of the book completes the answer a totally separate way.

    It takes the unified mass of N-13 and adds a neutron to it. It then takes this value and subtracts the unified mass of N-14 to get a difference of 0.0113 u. They convert this into an average binding value of 10.5 MeV/c2.

    They then go on to say it would require ΔE = Δmc2 = 10.5 MeV/c2 of energy to remove one neutron from the N-14 isotope.

    So my questions are as follows;
    1. Who is more correct for their average binding value the book or me?
    2. Why is my method wrong if the book is right?
    3. For either my value or the books value, the final answer would've required me to convert the MeV/c2 value into a kg equivalent, and then plugged it into the E=mc2 equation to get a proper Joules value.

    I appreciate any clarification you can offer.
     
  2. jcsd
  3. Jul 12, 2012 #2
    Unfortunately I don't quite follow what you are trying to do. It doesn't seem to make much sense to me :confused:

    Anyway though, I will explain the approach that the book takes. In fact, you don't have to concern yourself with the binding energy at all.
    The book uses the principle of conservation of energy to tackle the problem:

    Added Energy + mass-energy of N-14 = mass-energy of N-13 + mass-energy of a free neutron.

    So, Energy needed = (13.0057 u + 1.008665 u - 14.0031u) c2 = 0.0113 u c2
     
  4. Jul 12, 2012 #3
    Thanks Fightfish,

    So to clarify my previous thought process.

    In the book it starts off by saying take the unified mass from the table it provieds of 147N and subtract the unified mass of all the electrons.

    It than says to take the unified mass of the components for the nucleus and add them togeter.

    Notice they're not the same. Find the differance.

    This differance is the excess mass that is missing in the final product.

    This missing mass must be the energy needed to create the element.

    this excess unified mass can be converted into a mass [kg] or binding energy [MeV/c2.

    The binding energy turns out to be the total binding energy. It goes on to say that the minimum amount of energy required to break up a nucleus into its constituent nucleons is the total binding energy.

    It then says to find the Average binding energy take the total and divide it by the number of components in the nucleus. This tells you how tight each necleon is held in the nucleus. I noticed it doesn't say anything about this number being needed to break up any part of the nucleus.

    So this element is floating around and randomly reacts in the atmosphere (or somehow reacts) and decides to lose a Neutron.

    147N → 137N + 10n

    Now I see where they came up with using the conservation of energy equation.

    At this point in the book it hadn't touched reaction formulas. After seeing a reaction formula this process makes sense.

    So my question is as follows, if the unified mass for 137N wasn't given could one find an answer to the following question.

    How much energy would be required to remove one neutron from nitrogen-14 isotope, given its mass is 14.0031 u?
     
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