# Question #7e3ca

May 23, 2016

$\text{32 g}$

#### Explanation:

In order to be able to solve this problem, you need to know the nuclear half-life of potassium-42, $\text{^42"K}$, a radioactive isotope of potassium, which you can find listed here

The nuclear half-life tells you how much time is needed for half of the atoms present in a sample of a radioactive substance to undergo radioactive decay. Potassium-42 has a half-life of $\text{12.36}$ hours, which means that it takes $12.6$ hours for half of an initial sample to decay.

If you take ${A}_{0}$ to be the initial mass of potassium-42, $A$ the mass that remains after $62.0$ hours, and $n$ to be the number of half-lives that pass in this time period, you can say that

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

To find the value of $n$, simply divide the period of time given to you by the half-life of the isotope

$n = \left(62.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{hours"))))/(12.36 color(red)(cancel(color(black)("hours}}}}\right) = \frac{62}{12.36}$

Rearrange the above equation to solve for ${A}_{0}$, the initial mass of the potassium-42 sample

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

Since you know that you're left with $\text{1 g}$ of potassium-42 after the given period of time passes, you will have

${A}_{0} = \text{1 g" * 2^(62/12.36) = "32.36 g}$

Rounded to two sig figs, the answer will be

${A}_{0} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{32 g} \textcolor{w h i t e}{\frac{a}{a}} |}}}$