Find the half-life (in hours) of a radioactive substance that is reduced by 10 percent in 55 hours.

Answer 1

M = M₀e^rt; If t = 55 hours then M = 0.9M₀ and we have: 0.9M₀ = M₀e^55r; 0.9 = e^55t; r = ln0.9/55 =

- 0.0001898 h^-1. So, M = M₀e^-0.0001898. If t = T half-time time then M = 0.5M₀.

We have: 0.5M₀ = M₀e^-0.0001898T: T = ln0.5/(- 0.0001898) = 3652 hours