Question

use the exponential decay model, $a = a_0e^{kt}$, to solve the following. the half - life of a certain substance is 23 years. how long will it take for a sample of this substance to decay to 84% of its original amount? it will take approximately for the sample of the substance to decay to 84% 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 $t = 23$ (half - life), $A=\frac{1}{2}A_0$. Substitute into $A = A_0e^{kt}$:
$\frac{1}{2}A_0=A_0e^{23k}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $\frac{1}{2}=e^{23k}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{23k})$. Using the property $\ln(e^{x}) = x$, we have $\ln(\frac{1}{2}) = 23k$. So, $k=\frac{\ln(\frac{1}{2})}{23}\approx\frac{- 0.6931}{23}\approx - 0.0301$.

Step2: Find the time t when $A = 0.84A_0$

Substitute $A = 0.84A_0$ and $k\approx - 0.0301$ into $A = A_0e^{kt}$:
$0.84A_0=A_0e^{-0.0301t}$. Divide both sides by $A_0$ (since $A_0
eq0$), we get $0.84 = e^{-0.0301t}$. Take the natural logarithm of both sides: $\ln(0.84)=\ln(e^{-0.0301t})$. Using the property $\ln(e^{x}) = x$, we have $\ln(0.84)=-0.0301t$. Then $t=\frac{\ln(0.84)}{- 0.0301}\approx\frac{-0.1744}{-0.0301}\approx5.8$.

Answer:

$5.8$