## Related questions with answers

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

VerifiedUse 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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