Question #7bdb7

Oct 18, 2017

Here's what I got.

Explanation:

The idea here is that a radioactive isotope's nuclear half-life tells you the time needed for half of an initial sample to undergo radioactive decay.

In your case, you know that it takes $3$ minutes for half of any amount of polonium-218 that you have to undergo radioactive decay.

If you take ${A}_{0}$ to be the initial mass of polonium-238 and ${A}_{t}$ to be the amount that remains undecayed after a given period of time $t$, you can say that you have

${A}_{t} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{n}$

Here

• $n$ is the number of half-lives that pass in the given period of time $t$

In your case, you know that it takes $30$ minutes to transport the sample of polonium-238, which implies that you have

$n = \left(30 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{minutes"))))/(3color(red)(cancel(color(black)("minutes}}}}\right) = 10$

So if $10$ half-lives pass in $30$ minutes and you know that you must end up with $\text{0.10 g}$ of polonium-238, you can say that you have

$\text{0.10 g} = {A}_{0} \cdot {\left(\frac{1}{2}\right)}^{10}$

Rearrange to solve for ${A}_{0}$

${A}_{0} = \text{0.10 g" * 2^10 = "102.4 g}$

Now, I'll leave the answer rounded to two sig figs, but you could round it to one significant figure based on the value you have for the half-life of the isotope.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{amount needed" = 1.0 * 10^2color(white)(.)"g}}}}$