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

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

Step-by-step explanation:

The differential equation for decay is

`(dA)/(dt)= kA`

`\Rightarrow (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*3.3)`

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

`\Rightarrow k=(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)* t)`

`\Rightarrow e^(-0.21t)=0.2`

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

`\Rightarrow t= (ln (0.2))/(-0.21)`

⇒ t≈7.66

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