Question

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 80% of the lead to decay

Answer 1

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

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

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

Integrating both sides

ln A= kt+c₁

$\Rightarrow A= e^{kt+c_1}$

$\Rightarrow A=Ce^{kt}$         [$e^{c_1}=C$]

The initial condition is A(0)= A₀,

$\therefore A_0=Ce^{0.k}$

$\Rightarrow C=A_0$

$\therefore A=A_0e^{kt}$.........(1)

Given that the $Pb_{209}$  has half life of 3.3 hours.

For half life $A=\frac12 A_0$ putting this in equation (1)

$\frac12A_0=A_0e^{k\times3.3}$

$\Rightarrow ln(\frac12)= 3.3k$     [taking ln both sides, $ln \ e^a=a$]

$\Rightarrow k=\frac{ln \frac12}{3.3}$

⇒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)

$\therefore 0.2=1e^{(-0.21)\times t}$

$\Rightarrow e^{-0.21t}=0.2$

$\Rightarrow -0.21t= ln(0. 2)$

$\Rightarrow t= \frac{ln (0.2)}{-0.21}$

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.