Question

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?

Answer 1

`"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")`