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

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.