# Question e2e73

Apr 30, 2016

$\text{0.19 Ci}$

#### Explanation:

The idea here is that you need to find the amount of phosphorus-32 that will decay in five days to leave behind $\text{0.15 Ci}$.

Notice that problem provides you with the isotope's nuclear half-life, which as you know tells you how much time is needed for a sample of radioactive substance to decay to half of its initial value.

If you take ${A}_{0}$ to be the initial amount of phosphorus-32 and $A$ to be amount of phosphorus-32 that you need for the experiment, i.e. $\text{0.15 Ci}$, you can say that you have

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} A = {A}_{0} \cdot \frac{1}{2} ^ n \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here $n$ represents the number of half-lives that pass in a given amount of time and is calculated using

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} n = \text{period of time"/"half-life} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You know that the sample must travel for $5$ days, and that phosphorus-32 has a half-life of $14.3$ days, which means that $n$ will be equal to

$n = \left(5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{days"))))/(14.3color(red)(cancel(color(black)("days}}}}\right) = 0.34965035$

Rearrange the first equation to solve for ${A}_{0}$, the amount of phosphorus-32 that is needed in order to have $\text{0.15 Ci}$ left after $n$ half-lives pass

$A = {A}_{0} \cdot \frac{1}{2} ^ n \implies {A}_{0} = A \cdot {2}^{n}$

Plug in your values to find

A_0 = "0.15 Ci" * 2^0.34965035 = color(green)(|bar(ul(color(white)(a/a)"0.19 Ci"color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs.