Question

The mass of carbon in an animal bone fragment found in an archeology site is 200 g. If the bone registers an activity of 16 Bq, what is its age? Note: 12.0 g of C¹²₆ contains 6.02 x 1023 atoms and C¹⁴₆ / C¹²₆ ~ 1.3x10^-12

Answer 1

The activity of 16 decays (probably per second, since we're given the number of seconds in a year for this problem) means 16 atoms of carbon-14 decay every second in the sample.

With such a small fraction of the carbon atoms in a sample being C-14, we can consider the atomic weight of carbon in the sample to be 12. Each gram of carbon will contain 1/12 mole, or
1/12 * 6.02e23 atoms = 5.01e22 atoms/gram

In a fresh sample of carbon, we expect to see 5.01e22 * 1.3e-12 = 6.51e10 atoms of C-14 per gram.

Now, how many of those atoms will be decaying in any given second?

The half-life of C-14 is 5730 years.
The average lifetime of a C-14 atom is equal to the half-life divided by the natural log of 2.
5730/0.6931 = 8267 years. Since a year is 31.56 megaseconds, the average lifetime is
3.156e7 * 8.267e3 = 2.609e11 seconds.

The chance that any given atom of C-14 will decay during any given second is equal to one over the average lifetime. So the number of decays per second in a gram of fresh carbon would be:

A = 6.51e10 / 2.609e11 = 2.50e-1 = 0.25

So in a 200 gram sample, we expect to see 200 * 0.25 = 50 decays/second.

We actually see 16, so some of the original C-14 has decayed away. How much?

A(t) / A(0) = 2^(-t/h)

where A(t) is the amount at time t
t is the elapsed time
h is the half-life.

16/50 = 0.32 = 2^(-t/5730)

Take logs of both sides of the "=" sign:

-0.495 = -t/5730 * 0.301

-0.495 * 5730 / 0.301 = -t
t = 9423 years ==>ans

As a test of reasonableness, the amount of C-14 in the sample is 0.32 times the amount we'd expect in a fresh sample.
One half-life would result in the amount of C-14 decaying down to half the initial amount.
Two half-lives would result in the amount decaying down to one-quarter the initial amount.
So 0.32 should be somewhere between the two.
And it is.