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

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    $\begingroup$ Why don't you think you can use mass units for A? Do you mean mass excess? $\endgroup$
    – Bill N
    Mar 23, 2021 at 19:37

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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.

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According to Emilio Segre in his book Nuclei and Particles (pg. 190)

$A-M$ 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 $\Delta M = 0.$

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