Question #e2e73

1 Answer
Apr 30, 2016

Answer:

#"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 #"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. #"0.15 Ci"#, you can say that you have

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

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

#color(blue)(|bar(ul(color(white)(a/a)n = "period of time"/"half-life"color(white)(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 = (5 color(red)(cancel(color(black)("days"))))/(14.3color(red)(cancel(color(black)("days")))) = 0.34965035#

Rearrange the first equation to solve for #A_0#, the amount of phosphorus-32 that is needed in order to have #"0.15 Ci"# left after #n# half-lives pass

#A = A_0 * 1/2^n implies A_0 = A * 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.