Calculations Involving Binding Energy of Nucleus

[Lissajoux]

Homework Statement

The binding energy of a nucleus is given by:

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}}

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

\frac{dE_{b}}{dZ}=0

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

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

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

Homework Equations

Within the problem statement and subsequent solution attempt.

The Attempt at a Solution

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

\frac{dE_{b}}{dZ}=-2Za_{3}A^{-\frac{1}{3}}-a_{4}\left(2ZA^{-1}-1\right)=0

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

b) I know what the values of A, a_{3}, a_{4} 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 Z such that \frac{dE_{b}}{dZ}=0 right? Not sure how to do this.

c) Once I know what Z is, how do I determine which element of mass number 25 is a stable isotope? I don't really understand this question part.

 

[Lissajoux]

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 Z=12.

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

.. am I anywhere near with this?!