# What percent of a substance is left after six hours if a radioactive substance decays at a rate of 3.5% per hour?

Oct 26, 2014

Since the amount of the substance becomes 96.5% each hour, the amount $R \left(t\right)$ of a radioactive substance can be expressed as

$R \left(t\right) = {R}_{0} {\left(0.965\right)}^{t}$,

where ${R}_{0}$ is an initial amount, and $t$ is time in hours.

The percentage of the substance after 6 hours can be found by

$\frac{R \left(6\right)}{{R}_{0}} \cdot 100 = \frac{{R}_{0} {\left(0.965\right)}^{6}}{R} _ 0 \cdot 100 \approx 80.75$%

I hope that this was helpful.