Question

The core melt- down and explosions at the nuclear reactor in Chernobyl in 1986 released large amounts of strontium-90, which decays exponentially at the rate of 2.5% per year. Areas downwind of the reactor will be uninhabitable for 100 years. What percent of the original strontium-90 contamination will still be present after 50 years?

Answer 1

If
`S(t)` is the amount of strontium-90 present in the area in year
`t`, and it decays at a rate of 2.5% per year, then

`S(t+1)=(1-0.025)S(t)=0.975S(t)`

Let
`S(0)=s` be the starting amount immediately after the nuclear reactor explodes. Then

`S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^(t+1)S(0)`

or simply

`S(t)=0.975^ts`

So that after 50 years, the amount of strontium-90 that remains is approximately

`S(50)=0.975^(50)s\approx0.282s`

or about 28% of the original amount.

We can confirm this another way; recall the exponential decay formula,

`S(t)=se^(kt)`

where
`t` is measured in years. We're told that 2.5% of the starting amount
`s` decays after 1 year, so that

`0.975s=se^k\implies k=\ln0.975`

Then after 50 years, we have

`S(50)=se^(50k)\approx0.282s`