Question #7e3ca

1 Answer
May 23, 2016

Answer:

#"32 g"#

Explanation:

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

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.

http://www.pnausa.org/harmony-todd/nuclear-101-radioactive-half-life

Potassium-42 has a half-life of #"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

#color(blue)(|bar(ul(color(white)(a/a)A = A_0 * 1/2^ncolor(white)(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 = (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#, the initial mass of the potassium-42 sample

#A_0 = A * 2^n#

Since you know that you're left with #"1 g"# of potassium-42 after the given period of time passes, you will have

#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)|)))#