Question

use the exponential decay model, $a = a_0e^{kt}$, to solve the following. the half - life of a certain substance is 15 years. how long will it take for a sample of this substance to decay to 66% of its original amount? it will take approximately for the sample of the substance to decay to 66% of its original amount. (round the final answer to one decimal place as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Find the decay constant k

When $t = 15$ (half - life), $A=\frac{1}{2}A_0$. Substitute into $A = A_0e^{kt}$:
$\frac{1}{2}A_0=A_0e^{15k}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $\frac{1}{2}=e^{15k}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{15k})$. Using the property $\ln(e^x)=x$, we have $\ln(\frac{1}{2}) = 15k$. Then $k=\frac{\ln(\frac{1}{2})}{15}\approx\frac{- 0.6931}{15}\approx - 0.0462$.

Step2: Find the time t when $A = 0.66A_0$

Substitute $A = 0.66A_0$ and $k\approx - 0.0462$ into $A = A_0e^{kt}$:
$0.66A_0=A_0e^{-0.0462t}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $0.66 = e^{-0.0462t}$. Take the natural logarithm of both sides: $\ln(0.66)=\ln(e^{-0.0462t})$. Using the property $\ln(e^x)=x$, we have $\ln(0.66)=-0.0462t$. Then $t=\frac{\ln(0.66)}{- 0.0462}\approx\frac{-0.4155}{-0.0462}\approx9.0$.

Answer:

$9.0$