A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.

Solution:

Use the formula:

S=Ce^{-kt}

First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume that S = 0.5So when t = 38 hrs and C = So.

\left(0.5\right)So=\left(So\right)e^{-k\left(38\right)}

And then solve for k:

k=-0.018241

And then substitute k to the formula:

S=Ce^{-0.018241\left(t\right)}

Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.

\left(0.1\right)So=\left(So\right)e^{-0.018241\left(t\right)}

And then solve for time(t):

t\:=\:126.23\:hrs