Question

the decay rate, k, for a particular radioactive element is 18%, where time is measured in years. find the half - life of the element.
the half - life is
(round to one decimal place as needed )

Explanation:

Step1: Recall the decay formula

The general formula for exponential decay is $A = A_0e^{-kt}$, where $A$ is the amount of the substance at time $t$, $A_0$ is the initial amount, $k$ is the decay - rate constant, and $t$ is the time. For half - life, $A=\frac{A_0}{2}$.

Step2: Substitute into the decay formula

Substitute $A = \frac{A_0}{2}$ into $A = A_0e^{-kt}$. We get $\frac{A_0}{2}=A_0e^{-kt}$. Since $A_0
eq0$, we can divide both sides by $A_0$ to obtain $\frac{1}{2}=e^{-kt}$.

Step3: Solve for $t$

Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{-kt})$. Using the property $\ln(e^x)=x$, we have $\ln(\frac{1}{2})=-kt$. Given $k = 0.018$ (since $1.8\%=0.018$), then $t=-\frac{\ln(\frac{1}{2})}{k}$.

Step4: Calculate the value of $t$

We know that $\ln(\frac{1}{2})=-\ln(2)\approx - 0.693$. So $t=\frac{\ln(2)}{k}=\frac{0.693}{0.018}\approx38.5$.

Answer:

$38.5$