Question

# The following radioactive isotopes are used in medicine for imaging organs, studying blood circulation, treating cancer, and so on. Give the number of neutrons present in each isotope:$^{198}\mathrm{Au}$$^{47}\mathrm{Ca}$$^{60}\mathrm{Co}$$^{18}\mathrm{F}$$^{125}\mathrm{I}$$^{131}\mathrm{I}$$^{42}\mathrm{K}$$^{43}\mathrm{K}$$^{24}\mathrm{Na}$$^{32}\mathrm{P}$$^{85}\mathrm{Sr}$$^{99}\mathrm{Tc}$

Solution

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Use the periodic table to look up the atomic number for gold. Since the atomic number for gold is $Z=79$, then we know that a gold atom has 79 protons and 79 electrons.

Calculate the number of neutrons of gold-198 by subtracting its atomic number from its mass number.

\begin{align*} N&=A-Z\\ &=198-79\\ &=119 \end{align*}

Therefore, gold-198 has 119 neutrons.

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