The isotope 79Au198 (atomic mass 197.968 u) of gold, which has a half-life of 2.69 days, is used in cancer therapy. What mass (in grams) of this isotope is required to produce an activity of 218 Ci?
Added by Nicole L.
Step 1
First, we need to find the decay constant (λ) using the half-life (t½) formula: t½ = ln(2) / λ 2.69 days = ln(2) / λ Show more…
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$$ \text { The isotope }{ }_{79}^{198} \mathrm{Au} \text { (atomic mass }=197.968 \mathrm{u} \text { ) of gold has } $$ a half-life of 2.69 days and is used in cancer therapy. What mass (in grams) of this isotope is required to produce an activity of $315 \mathrm { }^{\circ}{C} ?$
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