# How many electrons, protons, and neutrons in the ""^67Zn isotope?

In ""^67Zn........I count $67 - 30 = 37$ $\text{neutrons}$.
Each zinc isotope contains 30 protons, 30 massive, positively charged nuclear particles. That $Z = 30$, $\text{(Z=the atomic number)}$, defines the nucleus as ZINC. And of course if there are 30 positively charged particles, there must be 30 negatively charged particles in the NEUTRAL atom. Agree?
But the atomic mass is to a first approx. the number of massive particles, neutrons, and protons. We gots 30 protons, and there must 37 neutrons to make up the mass number, and give the ""^67Zn isotope. From memory this isotope is approx. 5% abundant (but look it up!). The weighted average mass of all the isotopes is quoted as the atomic mass of zinc on the Periodic Table, $65.4 \cdot g \cdot m o {l}^{-} 1$.
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