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,
[tex]\frac{dA}{dt}\propto A[/tex]
[tex]\implies \frac{dA}{dt}=kA[/tex]
[tex]\int \frac{dA}{A}=\int kdt[/tex]
[tex]ln|A|=kt+C_1[/tex]
[tex]A=e^{kt+C_1}[/tex]
[tex]A=e^{C_1} e^{kt}[/tex]
[tex]\implies A=C e^{kt}[/tex]
Let [tex]A_0[/tex] be the initial amount,
[tex]A_0=C e^{0} = C[/tex]
[tex]\implies A=A_0 e^{kt}[/tex]
Since, the half-life of 3.3 hours.
[tex]\implies \frac{A_0}{2}=A_0 e^{3.3k}\implies e^{3.3k}=0.5\implies k=-0.21004[/tex]
[tex]\implies A=A_0 e^{-0.21004t}[/tex]
Here, [tex]A_0=1\text{ gram}[/tex]
[tex]A=(100-85)\% \text{ of }A_0=15\%\text{ of }A_0=0.15A_0[/tex]
By substituting the values,
[tex]0.15A_0=A_0 e^{-0.21004t}[/tex]
[tex]0.15=e^{-0.21004t}[/tex]
[tex]\implies t\approx 10.96\text{ hour}[/tex]
