Question

(1 point) find the half-life (in hours) of a radioactive substance that is reduced by 1010 percent in 9595 hours.

Answer 1

The decay function is of the form

`N(t) = N_(0) \, e^(-kt)`
where
N₀ = initial amount
k = decay constant
t = hours

The material decays by 10% in 95 hours. Therefore

`0.9N_(0) = N_(0) \, e^(-95k) \\\\ -95k = ln(0.9) \\\\ k= (ln(0.9))/(-95)=0.001109`

The time for the half life is given by

`0.5N_(0) = N_(0) \, e^(-0.001109t) \\\\ -0.0001109t = ln(0.5) \\\\ t = (ln(0.5))/(-0.001109) = 625 \, h`

Answer: The half life is 625 hours