# Question #7e3ca

##### 1 Answer

#### Answer:

#### Explanation:

In order to be able to solve this problem, you need to know the **nuclear half-life** of *potassium-42*,

https://en.wikipedia.org/wiki/Isotopes_of_potassium

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 **hours**, which means that it takes **half** of an initial sample to decay.

If you take **remains** after **hours**, and **number of half-lives** that pass in this time period, you can say that

#color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^ncolor(white)(a/a)|)))#

To find the value of *half-life* of the isotope

#n = (62.0 color(red)(cancel(color(black)("hours"))))/(12.36 color(red)(cancel(color(black)("hours")))) = 62/12.36#

Rearrange the above equation to solve for

#A_0 = A * 2^n#

Since you know that you're left with

#A_0 = "1 g" * 2^(62/12.36) = "32.36 g"#

Rounded to two **sig figs**, the answer will be

#A_0 = color(green)(|bar(ul(color(white)(a/a)"32 g"color(white)(a/a)|)))#