Find the half-life (in hours) of a radioactive substance that is reduced by 15 percent in 60 hours
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Originally Posted by cwarzecha Find the half-life (in hours) of a radioactive substance that is reduced by 15 percent in 60 hours Again $\displaystyle A(t) = A_0 \left ( \frac{1}{2} \right ) ^{t/t_{1/2}}$ So $\displaystyle \frac{A(t)}{A_0} = 0.15 = \left ( \frac{1}{2} \right ) ^{60/t_{1/2}}$ Solve for $\displaystyle t_{1/2}$. -Dan
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