If 2.5 grams of a radioactive isotope exist after 30 days in a storage vial, how much was present when it was sealed if the t_½ = 6 days? A)80 grams B)60 grams C)40 grams D)20 grams

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Quizplus · Accepted Answer

The amount of a radioactive isotope present after a certain time can be calculated using the formula $A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$, where $A$ is the amount after time $t$, $A_0$ is the initial amount, and $t_{1/2}$ is the half-life of the isotope. Given $A = 2.5$ grams, $t = 30$ days, and $t_{1/2} = 6$ days, we can solve for $A_0$. Rearranging the formula gives $A_0 = A \left(\frac{1}{2}\right)^{-\frac{t}{t_{1/2}}}$. Plugging in the given values, $A_0 = 2.5 \left(\frac{1}{2}\right)^{-\frac{30}{6}} = 2.5 \left(\frac{1}{2}\right)^{-5} = 2.5 \times 32 = 80$ grams.

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