Question

The amount of a radioactive isotope present in a certain sample at time t is given by A(t)=600e−0.02838t grams, where t years is the time since the initial amount was measured.a. Find the initial amount of the isotope that is present in the sample.b. Find the half-life of this isotope. That is, find the number of years until half of the original amount of the isotope remains.

Answer 1

We have the following equation for the amount in grams of radiactive isotope:

$A(t)=600\cdot e^{-0.02828t}$

where t denotes the time in years.

a. Find the initial amount of the isotope that is present in the sample.

The initial amount of the isotope occurs at t=0. Then, by substituting this value into to our equation, we have

$\begin{gathered} A(0)=600\cdot e^{0.02828(0)} \\ A(0)=600\cdot1 \\ A(0)=600 \end{gathered}$

Then, the initial amount is 600 grams.

b. Find the​ half-life of this isotope.

In this case, the final amount must be half of the starting material, that is,

$A(t)=\frac{600}{2}=300$

and the half-life is the corresponding time for this amount of material. So, we have

$300=600\cdot e^{-0.02838\cdot t}$

By moving 600 to the left hand side, we get

$\begin{gathered} \frac{300}{600}=e^{-0.02838\cdot t} \\ or\text{ equivalently.} \\ e^{-0.02838\cdot t}=\frac{1}{2} \end{gathered}$

therefore, the time is given as

$t=-\frac{1}{0.02828}\ln (\frac{1}{2})$

then, the time is

$t=24.423\text{ years}$

then. by rounding down, the answer is 24 years.