The half-life of phosphorus-32 is 14.29 days and decays according to the formula $F=A e^{-0.0485 t}$, where $F$ is the final amount, $A$ is the initial amount and $t$ is the time in days. How much phosphorus- 32 is left after 60 days, if the initial amount was $25.0 \mathrm{mg}$ ?

Question

Quizplus · Accepted Answer

The Answer of The half-life of phosphorus-32 is 14.29 days and decays according to the formula $F=A e^{-0.0485 t}$, where $F$ is the final amount, $A$ is the initial amount and $t$ is the time in days. How much phosphorus- 32 is left after 60 days, if the initial amount was $25.0 \mathrm{mg}$ ?

The Half-Life of Phosphorus-32 Is 14 F=Ae−0.0485tF=A e^{-0.0485 t}F=Ae−0.0485t , Where FFF

The Half-Life of Phosphorus-32 Is 14 $F=A e^{-0.0485 t}$ , Where $F$