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Homework Help: Calculations Involving Binding Energy of Nucleus

  1. Apr 24, 2010 #1
    1. The problem statement, all variables and given/known data

    The binding energy of a nucleus is given by:

    [tex]E_{b}=a_{1}A-a_{2}A^{\frac{2}{3}}-a_{3}Z^{2}A^{-\frac{1}{3}}-a_{4}\left(Z-\frac{A}{2}\right)^{2}A^{-1}\pm a_{5}A^{-\frac{1}{2}}[/tex]

    For a given set of isobars, [itex]A[/itex] constant, [itex]E_{b}[/itex] will be maximised at the value of [itex]Z[/itex] that satisfies:


    a) Find this derivative and solve for [itex]Z[/itex] in terms of the [itex]a_{i}[/itex] and [itex]A[/itex]

    b) Using the expression derived in part a) and given values of [itex]a_{i}[/itex], find the value of [itex]Z[/itex] which maximises [itex]E_{b}[/itex] when [itex]A=25[/itex]. Round [itex]Z[/itex] to the nearest integer value.

    c) Use the periodic table and the derived value of [itex]Z[/itex] from part b) to determine which element of mass number 25 is a stable isotope.

    2. Relevant equations

    Within the problem statement and subsequent solution attempt.

    3. The attempt at a solution

    a) This is what I have for the differentiated equation:


    .. this is correct? I think that's all I have to do for this part.

    b) I know what the values of [itex]A, a_{3}, a_{4}[/itex] are, so I can put these into the equation, but then how to I 'solve it for Z' from that? I guess that need to find a value of [itex]Z[/itex] such that [itex]\frac{dE_{b}}{dZ}=0[/itex] right? Not sure how to do this.

    c) Once I know what [itex]Z[/itex] is, how do I determine which element of mass number 25 is a stable isotope? I don't really understand this question part.
    Last edited: Apr 25, 2010
  2. jcsd
  3. Apr 24, 2010 #2
    Just gone back and looked at this again.

    Using the differential equation calculated, as listed above, I've plugged in the values I know and solved for Z. Have got an answer of [itex]Z=12[/itex].

    So this I assume means I'm looking for whatever is [itex]{}_{12}^{25}?[/itex] which by looking at the periodic table I find that this is [itex]{}_{12}^{25}Mn[/itex] where the 12 just means that's the stable isotope.

    .. am I anywhere near with this?!
    Last edited: Apr 25, 2010
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