First we will assume this is a first order nuclear reaction, in which case we can use two equations. The first equation will help us find our decay constant, k:
t1/2 = 0.693/k
This helps us find the decay constant using the known half life (which you gave in the question). Plug in the half life of 36 days and we should get a decay constant k of 0.01925. Knowing this, we can use the equation for first order nuclear decay:
ln(N/No) = -kt
where N is the current (or in this case, desired) amount of nuclear material, No is the initial amount of material, and k is the decay constant calculated above. We will be solving for t, the time. Plugging in the values from the problem, the equation should look like:
ln(100/2000) = -(0.01925)t
Plugging into our calculator, we get t=155.6. Our units are in days since we plugged 36 days into our first equation to solve for k. This means it will take approximately 156 days for the lake to be safe to swim in again.
We can conceptually check this number by going through chunks of 36 days and estimating the amount of material. After 36 days, there will be 1000 tons; after 72 days, 500 tons; after 108 days, 250 tons; after 144 days, 125 tons; after 180 days, 62.5 tons. It's logical that our desired tonnage would occur between 144 and 180 days.
Julia S.
This equation is still great when you know t and t(1/2)! I tried rewriting your final equation as an expression of logs, but it didn’t work. :-(07/21/21