Carbon-14 has a half-life of 5770 years. If a fossil is 23,080 years old and it has 3 Kg of Carbon-14, how much carbon-14 did it originally have?

1 Answer
Jun 21, 2016

"48 kg"48 kg

Explanation:

The nuclear half-life is simply the time needed for half of the atoms present in a sample of a radioactive substance to undergo radioactive decay.

![http://www.pnausa.org/harmony-todd/nuclear-101-radioactive-half-life](useruploads.socratic.org)

In your case, carbon-14 is said to have a half-life of 57705770 years, which means that the mass of any sample of carbon-14 will be halved with the passing of 57705770 years.

Now, if 11 half-life from now the mass of the sample will be halved, it follows that 11 half-life ago it was twice its current value.

An interesting technique to use here is to backtrack from current time, when the sample is down to "3 kg"3 kg, to "23,080"23,080 years ago.

You can thus say that 57705770 years before this point, the mass of the sample was twice the current mass

2 xx "3 kg" = "6 kg"2×3 kg=6 kg

How about 22 half-lives ago?

2 xx "5770 years" = "11,540 years ago"2×5770 years=11,540 years ago

The mass of the sample was

2 xx "6 kg" = "12 kg"2×6 kg=12 kg

How about 33 half-lives ago?

3 xx "5770 years" = "17,310 years ago"3×5770 years=17,310 years ago

The mass of the sample was

2 xx "12 kg" = "24 kg"2×12 kg=24 kg

Finally, how about 44 half-lives ago?

4 xx "5770 years" = "23,080 years ago"4×5770 years=23,080 years ago

The mass of the sample was

2 xx "24 kg" = "48 kg"2×24 kg=48 kg

Therefore, "23,08023,080 years ago, the sample contained "48 kg"48 kg of carbon-14.

When the values don't allow for such simple calculations, you can use the equation

color(blue)(|bar(ul(color(white)(a/a)"A"_t = "A"_0 * 1/2^ncolor(white)(a/a)|)))

Here

"A"_t - the mass of the sample that remains after a period of time t
"A"_0 - the initial mass of the sample
n - the number of half-lives that pass in the period of time t

In your case, you have

n = ("23,080" color(red)(cancel(color(black)("years"))))/(5770color(red)(cancel(color(black)("years")))) = 4

You will thus have

"A"_0 = "A"_t * 2^4

"A"_0 = "3 kg" * 2^4 = color(green)(|bar(ul(color(white)(a/a)color(black)("48 kg")color(white)(a/a)|)))