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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:

198Au^{198}\mathrm{Au}

47Ca^{47}\mathrm{Ca}

60Co^{60}\mathrm{Co}

18F^{18}\mathrm{F}

125I^{125}\mathrm{I}

131I^{131}\mathrm{I}

42K^{42}\mathrm{K}

43K^{43}\mathrm{K}

24Na^{24}\mathrm{Na}

32P^{32}\mathrm{P}

85Sr^{85}\mathrm{Sr}

99Tc^{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=79Z=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.

N=AZ=19879=119\begin{align*} N&=A-Z\\ &=198-79\\ &=119 \end{align*}

Therefore, gold-198 has 119 neutrons.

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