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

$S(50)=se^{50k}\approx0.282s$