Question

My Notes OAsk Your Tea The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 85% of the lead to decay? (Round your answel to two decimal places.) hr

Answer 1

Answer:

10.96 hours will take for 85% of the lead to decay.

Step-by-step explanation:

Suppose A represents the amount of Pb-209 at time t,

According to the question,

$\frac{dA}{dt}\propto A$

$\implies \frac{dA}{dt}=kA$

$\int \frac{dA}{A}=\int kdt$

$ln|A|=kt+C_1$

$A=e^{kt+C_1}$

$A=e^{C_1} e^{kt}$

$\implies A=C e^{kt}$

Let $A_0$ be the initial amount,

$A_0=C e^{0} = C$

$\implies A=A_0 e^{kt}$

Since, the half-life of 3.3 hours.

$\implies \frac{A_0}{2}=A_0 e^{3.3k}\implies e^{3.3k}=0.5\implies k=-0.21004$

$\implies A=A_0 e^{-0.21004t}$

Here, $A_0=1\text{ gram}$

$A=(100-85)\% \text{ of }A_0=15\%\text{ of }A_0=0.15A_0$

By substituting the values,

$0.15A_0=A_0 e^{-0.21004t}$

$0.15=e^{-0.21004t}$

$\implies t\approx 10.96\text{ hour}$