# Calculations Involving Binding Energy of Nucleus

1. Apr 24, 2010

### Lissajoux

1. The problem statement, all variables and given/known data

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.

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:

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

.. 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.

Last edited: Apr 25, 2010
2. Apr 24, 2010

### 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?!

Last edited: Apr 25, 2010