Question

the mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 6.8% per day. find the half - life of this substance (that is, the time it takes for one - half the original amount in a given sample of this substance to decay). note: this is a continuous exponential decay model. do not round any intermediate computations, and round your answer to the nearest hundredth. days

Explanation:

Step1: Recall the formula for half - life in continuous exponential decay

The formula for continuous exponential decay is $A = A_0e^{-rt}$, where $A$ is the amount of the substance at time $t$, $A_0$ is the initial amount, $r$ is the decay rate, and $t$ is the time. At half - life $t = t_{1/2}$, $A=\frac{A_0}{2}$. So, $\frac{A_0}{2}=A_0e^{-rt_{1/2}}$.

Step2: Simplify the equation

Divide both sides of the equation $\frac{A_0}{2}=A_0e^{-rt_{1/2}}$ by $A_0$ (since $A_0
eq0$), we get $\frac{1}{2}=e^{-rt_{1/2}}$.

Step3: Take the natural logarithm of both sides

$\ln(\frac{1}{2})=\ln(e^{-rt_{1/2}})$. Since $\ln(e^x)=x$, we have $\ln(\frac{1}{2})=-rt_{1/2}$. And $\ln(\frac{1}{2})=-\ln(2)$. So, $-\ln(2)=-rt_{1/2}$, and $t_{1/2}=\frac{\ln(2)}{r}$.

Step4: Substitute the given decay rate

Given $r = 0.068$. Then $t_{1/2}=\frac{\ln(2)}{0.068}$. Calculate $\ln(2)\approx0.693147$. So, $t_{1/2}=\frac{0.693147}{0.068}\approx10.19$.

Answer:

$10.19$