Semi-empirical mass formula [closed]

The mass formula is given by

$M(Z,A) = ZM_{p}+(A-Z)M_{n}-a_{1}A+a_{2}A^\frac{2}{3}+a_{3}\frac{Z(Z-1)}{A^\frac{1}{3}}+a_{4}\frac{(Z-A/2)^2}{A}+a_{5}A^\frac{-1}{2}$

So I am just wondering here what the $M(Z,A)$ stands for. Is it the actual mass of the atom which is obtained empirically or is it the sum of the masses of the constituent particles of the atom(i.e. electrons, protons, neutrons)? And if so, how would we use this to calculate the binding energy? I though the binding energy is equal to the mass deficit times $c^2$ which is why I want to know what $M(Z,A)$ stands for and how one would then use that information to calculate the binding energy of the nucleus

closed as off-topic by ACuriousMind♦Mar 15 '17 at 0:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you looked up what any of the other symbols in the equation mean? I suspect that if you look up what $ZM_p$ and $(A-Z)M_n$ the answer will become rather obvious – By Symmetry Apr 15 '15 at 15:39

So binding energy is: $B = |M_{atom}-M_{constituents}| = |M(Z,A) - ZM_{p}-(A-Z)M_{n}|\\ =|-a_{1}A+a_{2}A^\frac{2}{3}+a_{3}\frac{Z(Z-1)}{A^\frac{1}{3}}+a_{4}\frac{(Z-A/2)^2}{A}+a_{5}A^\frac{-1}{2}|$