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3 years ago

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

1 answer:

8 0

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
$0.9N_{0} = N_{0} \, e^{-95k} \\\\ -95k = ln(0.9) \\\\ k= \frac{ln(0.9)}{-95}=0.001109$

The time for the half life is given by
$0.5N_{0} = N_{0} \, e^{-0.001109t} \\\\ -0.0001109t = ln(0.5) \\\\ t = \frac{ln(0.5)}{-0.001109} = 625 \, h$

Answer: The half life is 625 hours

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