# 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?

Oct 19, 2015

$\text{2.28 g } \to$ after $1.00$ year
$\text{0.703 g } \to$ 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 $\text{5.30}$ 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

• ${A}_{0} / 2 \to$ after one half-life passes;
• ${A}_{0} / 2 \cdot \frac{1}{2} = {A}_{0} / 4 \to$ after two half-lives pass;
• ${A}_{0} / 4 \cdot \frac{1}{2} = {A}_{0} / 8 \to$ after three half-lives pass;
• ${A}_{0} / 8 \cdot \frac{1}{2} = {A}_{0} / 16 \to$ 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

$\text{remaining amount" = "initial amount"/2^n" }$, where

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

In your case, you know that you start with $\text{2.60 g}$ 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")