Semi empirical mass formula derivation help

[artis]

Homework Statement

None

Relevant Equations

$$BE = a_{v}A −a_{s}A^{2/3} −a_{c}\frac{Z(Z-1)}{A^{1/3}} −a_{A}\frac{(N-Z)^{2}}{A} + a_{p}\frac{1}{A^{1/2}}$$

Can somebody please derive for me an example of the Binding energy from the Semi Empirical mass formula? I am trying myself but always there is a difference between the database binding energy and my own result. I am calculating the BE of Niobium 93. For the mass formula I used the coefficients found on the table in the Wikipedia article, found here in the link
https://en.wikipedia.org/wiki/Semi-empirical_mass_formula

I used the coefficients in amu from the first graph by Eisberg & Resnick

Also , what is the purpose of the semi empirical mass formula, because once we know the masses of proton and neutron and the mass of each atom we can then use a simple adding and subtraction to find the binding energy of a particular nucleus which manifests itself as the mass defect.
Was it made to approximate the masses before we had the technical means to observe them empirically?

PS. This is not a typical homework , just me going through some relevant maths. I apologize for not being able to present the equation in LaTex , I tried but I failed.

 

[BvU]

I tried but I failed.

How can we help ?
(There's a good tutorial on LaTeX here.)

Enclose in double dollar signs to get displayed math:

$$ BE = avA −asA2/3 −acZ(Z −1)/A1/3 −aa(N −Z)/2A + ap1A/1/2 $$ gives

$$ BE = avA −asA2/3 −acZ(Z −1)/A1/3 −aa(N −Z)/2A + ap1A/1/2 $$ which is completely confusing when compared to the WIki picture

for which the input could be

$$ E_B = a_V A - a_S A^{2/3} - a_C { Z(Z-1)\over Z^{1/3} } - a_Z { (N-Z)^2 \over A } + \delta (N,Z) $$
which gives

$$ E_B = a_V A - a_S A^{2/3} - a_C { Z(Z-1)\over Z^{1/3} } - a_Z { (N-Z)^2 \over A } + \delta (N,Z) $$

##\ ##

 

[artis]

Ok, I tried the online latex editor and copied my result now using the dollar signs and it seems to work
thanks @BvU , but as I try to copy the same formula back in my OP under category "relevant equations" it doesn't display a nice formula anyway.

$$BE = a_{v}A −a_{s}A^{2/3} −a_{c}\frac{Z(Z-1)}{A^{1/3}} −a_{A}\frac{(N-Z)^{2}}{A} + a_{p}\frac{1}{A^{1/2}}$$Still I hope someone can show me a good derivation of this using what I explained in my OP #1 post

 

[jtbell]

always there is a difference between the database binding energy and my own result.

How big a difference? 50%? 0.5%?

As I understand it, the semi-empirical binding energy formula basically fits a "smooth" curve based on physical principles, to empirical data. Or, to put it another way, it tries to make some sense of observed measurements of nuclear masses, in terms of those physical principles. But it can only explain general trends.

Nuclei are complex systems, especially the larger ones. They're dominated by the "strong force" between fundamental particles, which is more complicated than the electromagnetic interaction. Some aspects are still not completely understood from first princples, out to umpteen decimal places, as far as I know.

 

[artis]

@jtbell well the difference was too large , don't remember exactly but since I calculated for Niobium 93 which is not one of those weird "off the charts" nuclei then I believe I got something wrong. I will try once more anyway.