$1.0 g$ of $_{79}^{198} A u$ $(t_{1 / 2} = 65 h o u r s)$ by $β$ -emission to produce mercury.
(a) Write the nuclear reaction for the process.
(b) How much mercury will be present after $260$ hours?

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Solution

(a) $_{79}^{198} A u ⟶_{80}^{198} H g +_{- 1}^{0} e$
(b) Number of half lives in $260$ hours $= \frac{260}{65} = 4$
Amount of gold left after $4$ half lives $= {(\frac{1}{2})}^{4} = \frac{1}{16} g$
Amount of gold disintegrated $= 1 - \frac{1}{16} = \frac{15}{16} g$
Amount of mercury formed $= \frac{15}{16} = 0.9375 g$

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