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 95% of the lead to decay?

Answer 1

`N(t) =N_o ((1)/(2))^{(t)/(t_(1/2))}`

Where
`t_(1/2)= 3.3 hr` represent the half life and the intial amount would be
`N_o = 1`

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

`0.05 = 1 (0.5)^(t/3.3)`

If we apply natural log on both sides we got:

`ln(0.05) = (t)/(3.3) ln (0.5)`

And solving for t we got:

`t= 3.3 *(ln(0.05))/(ln(0.5))= 14.26`

So then would takes about 14.26 hours in order to have 95% of the lead to decay

Explanation:

For this case we can define the variable of interest amount of Pb209 and for the half life would be given:

`N(t) =N_o ((1)/(2))^{(t)/(t_(1/2))}`

Where
`t_(1/2)= 3.3 hr` represent the half life and the intial amount would be
`N_o = 1`

And we want to find the time in order to have a 95% of decay so we can set up the following equation:

`0.05 = 1 (0.5)^(t/3.3)`

If we apply natural log on both sides we got:

`ln(0.05) = (t)/(3.3) ln (0.5)`

And solving for t we got:

`t= 3.3 *(ln(0.05))/(ln(0.5))= 14.26`

So then would takes about 14.26 hours in order to have 95% of the lead to decay