Question

Find the half-life of a radioactive element, which decays according to the function A(t) = Ao exp( -0.0294t), where t is the timein years.

Answer 1

Answer:

23.58 years.

Explanation:

The decay equation for the radioactive element is:

$A(t)=A_oe^{-0.0294t}\text{ where }\begin{cases}A(t)=\text{Amount at time t} \\ A_o=\text{Initial Amount}\end{cases}$

We want to find the half-life of the element.

The half-life of the radioactive substance is the time it will take for half of the initial amount of substance to decay. That is when:

$\begin{gathered} \text{Present Amount=}\frac{1}{2}\text{ of Initial Amount} \\ \implies A(t)=\frac{1}{2}A_o \end{gathered}$

Substitute A(t) into the formula.

$0.5A_o=A_oe^{-0.0294t}$

We then solve for t.

$\begin{gathered} \text{Divide both sides by 0.5} \\ \frac{0.5A_o}{A_o}=\frac{A_oe^{-0.0294t}}{A_o} \\ e^{-0.0294t}=0.5 \\ \text{Take the natural logarithm of both sides} \\ \ln (e^{-0.0294t})=\ln (0.5) \\ -0.0294t=\ln (0.5) \\ \text{Divide both sides by }-0.0294 \\ \frac{-0.0294t}{-0.0294}=\frac{\ln(0.5)}{-0.0294} \\ t=23.58\text{ years} \end{gathered}$

The half-life of the radioactive element is 23.58 years (correct to the nearest hundredth).