# How is Mass Defect dimensionally correct? [closed]

Where the Mass Defect is $$∆M = M-A$$. So how does $$(M-A)$$ have the unit of mass, as A is the mass number?

• Why don't you think you can use mass units for A? Do you mean mass excess? Commented Mar 23, 2021 at 19:37

The cute thing about atomic mass units is that, neglecting the proton-neutron mass difference and the binding energy, the mass number is the mass in amu. There aren't any isotopes with a mass excess or mass deficit greater than about $$100\,\mathrm{MeV}/c^2 \approx 0.1\,\mathrm{u}$$. If I tell you that this atom is tin-100 and you say "I bet the mass is 100 u," you are correct to four or five significant figures: the actual correction $$\Delta M$$ for tin-100 is about 60 MeV.
$$A-M$$ where $$M$$ is the exact mass of the atom, is usually called the "mass defect," the quantity $$M-A=\Delta M$$ is usually called the "mass excess"...
Both $$M$$ and $$A$$ are expressed in atomic mass units, u, where the neutral carbon-12 isotope has a mass of exactly 12.0000 u, so $$\Delta M = 0.$$