Question

Question #58689

Answer 1

`"7.5 g"`

Explanation:

The nuclear half-life of a given radioactive substance tells you how much time is needed until half of the atoms present in the sample undergo radioactive decay.

This means that if you start with an initial sample `A_0`, you can say that you will be left with

`A_0 * 1/2 = A_0/2 ->` after the passing of one half-life

`A_0/2 * 1/2 = A_0/4 ->` after the passing of two half-lives

`A_0/4 * 1/2 = A_0/8 ->` after the passing of three half-lives

`A_0/8 * 1/2 = A_0/16 ->` after the passing of four half-lives
`vdots`

and so on. Since the mass of the sample gets halved with every passing half-life, 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

`A` - the amount that remains after the passing of a given period of time
`A_0` - the initial mass of the sample
`n` - the number of half-lives that pass in a given period of time

You can find `n` by dividing the time that passed by the half-life of the sample.

`color(blue)(|bar(ul(color(white)(a/a)n = "how much time passed"/"half-life"color(white)(a/a)|)))`

In your case, you're interested in finding out how much radioactive material remains after the passing of `6000` years, so `n` will be equal to

`n = (6000 color(red)(cancel(color(black)("years"))))/(2000color(red)(cancel(color(black)("years")))) = 3`

So, after the passing of `3` half-lives, you will be left with

`A = "60 g" * 1/2^3 = color(green)(|bar(ul(color(white)(a/a)"7.5 g"color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.