# 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

#### Answer:

#### Explanation:

So, you're dealing with a sample of *cobalt-60*. You know that cobalt-60 has a nuclear half-life of

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_0/2 -># afterone half-lifepasses;#A_0/2 * 1/2 = A_0/4 -># aftertwo half-livespass;#A_0/4 * 1/2 = A_0/8 -># afterthree half-livespass;#A_0/8 * 1/2 = A_0/16 -># afterfour half-livespass;

#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

In your case, you know that you start with

How many cobalt-60 half-lives will pass in

#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

#10.0color(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")#