Question

A substance with a half life is decaying exponentially. If there are initially 12 grams of the substance and after 70 minutes there are 7 grams, after how many minutes will there be 2 grams remaining? Round your answer to the nearest whole number.

Answer 1

Answer: 233 min

Step-by-step explanation:

This problem can be solved by the following equation:

$A=A_{o} e^{-kt}$  (1)

Where:

$A=7 g$ is the quantity left after time $t$

$A_{o}=12 g$ is the initial quantity

$t=70 min$ is the time elapsed

$k$ is the constant of decay for the material

So, firstly we need to find the value of $k$ from (1) in order to move to the next part of the problem:

$\frac{A}{A_{o}}=e^{-kt}$  (2)

Applying natural logarithm on both sides of the equation:

$ln(\frac{A}{A_{o}})=ln(e^{-kt})$  (3)

$ln(\frac{A}{A_{o}})=-kt$  (4)

$k=-\frac{ln(\frac{A}{A_{o}})}{t}$  (5)

$k=-\frac{ln(\frac{7 g}{12 g})}{70 min}$  (6)

$k=0.00769995 min^{-1}$  (7)  Now that we have the value of $k$ we can solve the other part of this problem: Find the time $t$ for $A=2 g$.

In this case we need to isolate $t$ from (1):

$t=-\frac{ln(\frac{A}{A_{o}})}{k}$  (8)

$t=-\frac{ln(\frac{2 g}{12 g})}{0.00769995 min^{-1}}$  (9)

Finally:

$t=232.697 min \approx 233 min$