Question

If 14 g of a radioactive substance are present initially and 7 yr later only 7 g​ remain, how much of the substance will be present after 17 ​yr?

Answer 1

Answer:

2.60 gr

Step-by-step explanation:

We need to consider the function $P(t) = P_oe^{kt}$ where Po is the initial substance k is the rate of decay and t is the time

We know that $P_o = 14$ and at the 7 th year P(t) is 7

This means

$7 = 14e^{7k}$

We solve for k

$\frac{7}{14} = e^{7k}\\\\\ln (\frac{7}{14}) = 7k\\\\k = \frac{\ln (7/14)}{7} = -0.099\\\\Then \ P(t) = 14e^{-0.099t}\\\\Now \ we \ take \ t = 17\\\\P(17) = 14e^{(-0.099)(17)} = 2.60$