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

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

Explanation:

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

According to the question,

`(dA)/(dt)\propto A`

`\implies (dA)/(dt)=kA`

`\int (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 (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}`