Respuesta :
Answer:
Therefore it will take 7.66 hours for 80% of the lead decay.
Explanation:
The differential equation for decay is
[tex]\frac{dA}{dt}= kA[/tex]
[tex]\Rightarrow \frac{dA}{A}=kdt[/tex]
Integrating both sides
ln A= kt+c₁
[tex]\Rightarrow A= e^{kt+c_1}[/tex]
[tex]\Rightarrow A=Ce^{kt}[/tex] [[tex]e^{c_1}=C[/tex]]
The initial condition is A(0)= A₀,
[tex]\therefore A_0=Ce^{0.k}[/tex]
[tex]\Rightarrow C=A_0[/tex]
[tex]\therefore A=A_0e^{kt}[/tex].........(1)
Given that the [tex]Pb_{209}[/tex] has half life of 3.3 hours.
For half life [tex]A=\frac12 A_0[/tex] putting this in equation (1)
[tex]\frac12A_0=A_0e^{k\times3.3}[/tex]
[tex]\Rightarrow ln(\frac12)= 3.3k[/tex] [taking ln both sides, [tex]ln \ e^a=a[/tex]]
[tex]\Rightarrow k=\frac{ln \frac12}{3.3}[/tex]
⇒k= - 0.21
Now A₀= 1 gram, 80%=0.8
and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram
Now putting the value of k,A and A₀in the equation (1)
[tex]\therefore 0.2=1e^{(-0.21)\times t}[/tex]
[tex]\Rightarrow e^{-0.21t}=0.2[/tex]
[tex]\Rightarrow -0.21t= ln(0. 2)[/tex]
[tex]\Rightarrow t= \frac{ln (0.2)}{-0.21}[/tex]
⇒ t≈7.66
Therefore it will take 7.66 hours for 80% of the lead decay.
