If 2.60 g of Cobalt-60 (half-life = 5.30 y) are allowed to decay, how many grams would be left after 1.00 y and after 10.0 y?

1 Answer
Oct 19, 2015

Answer:

#"2.28 g " -># after #1.00# year
#"0.703 g " -># after #10.0# years

Explanation:

So, you're dealing with a sample of cobalt-60. You know that cobalt-60 has a nuclear half-life of #"5.30"# years, and are interested in finding how many grams of the sample would remain after #1.00# year and #10.0# years, respectively.

A radioactive isotope's half-life tells you how much time is needed for an initial sample to be halved.

If you start with an initial sample #A_0#, then you can say that you will be left with

  • #A_0/2 -># after one half-life passes;
  • #A_0/2 * 1/2 = A_0/4 -># after two half-lives pass;
  • #A_0/4 * 1/2 = A_0/8 -># after three half-lives pass;
  • #A_0/8 * 1/2 = A_0/16 -># after four half-lives pass;
    #vdots#

and so on.

Notice that you can find a relationship between the amount of a sample that remains and the number of half-lives that pass

#"remaining amount" = "initial amount"/2^n" "#, where

#n# - the number of half-lives that pass.

In your case, you know that you start with #"2.60 g"# of cobalt-60 and that you're interested in finding out how much you'll be left with after #1.00# year.

How many cobalt-60 half-lives will pass in #1.00# year?

#1.00color(red)(cancel(color(black)("year"))) * "1 half-life"/(5.30color(red)(cancel(color(black)("years")))) = "0.18868"#

This means that you'll be left with

#m_"1.00 year" = "2.60 g"/2^0.18868 = color(green)("2.28 g")#

How about after #10.0# years pass?

#10.0color(red)(cancel(color(black)("year"))) * "1 half-life"/(5.30color(red)(cancel(color(black)("years")))) = "1.8868"#

This time you'll be left with

#m_"10.0 years" = "2.60 g"/2^1.8868 = color(green)("0.703 g")#