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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

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

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    $\begingroup$ 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 $\endgroup$ – By Symmetry Apr 15 '15 at 15:39
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Its the mass of the atom with atomic number Z and mass number A. If you study the right side of the equation you will see that the first two terms are the total mass of the protons and the total mass of the neutrons. All other terms are modifying this sum of the masses of the constituent particles. So to calculate the mass defect, simply take the first two terms on the right to the left side.

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

A complete description can be found in Eisberg and Resnik.

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See the Wikipedia article on the semi-empirical mass formula. The formula is for the mass of the nucleus, so you would need to add on the mass of the electrons (and subtract their binding energy) to get the mass of the atom.

However, as previous questions have mentioned, the mass of the electrons is a small correction and in fact generally smaller than the errors in the semi-empirical mass formula.

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