Half-Life #9

Question

Radioactive Carbon-14 has a half-life of approximately 5730 years. It collects in living plants and plant-eating organisms and will start to decay once the organism dies.
Assume an animal bone is discovered in an archeological dig and is Carbon-14 dated to be roughly 6000 years old.
What fraction of the initial amount of Carbon-14 is remaining?

(Hint: consider the initial amount of Carbon-14 as 100% or just the number 1 and round to three decimals places. For example: 43.57% = 0.436 for the actual answer.)

Answer 1

There's a specific formula for finding the fraction left after a specific amount of time:

 

F = 0.5^(t/h), where F is the fraction left, t is the amount of time, and h is the half-life.

 

If you don't know that formula, you can still think it through using the general formula for growth or decay, and come up with the right answer.

 

The idea is that you want to find the rate of decay from the half-life and apply it to the period of 6000 years.  One thing you know immediately is that since 6000 years is slightly more than the half-life, then there should be slightly less than 1/2 left.  You can use fact that to check the answer at the end.

 

So here's one way to look at it.

 

Suppose you define a unit of time call the splurg, which is equal to 5730 years.  Because the splurg is equal to the half-life, that means that half the carbon-14 has decayed after 1 splurg, or that the rate of decay is 50% per splurg.

 

So how many splurgs are there in 6000 years?  That's easy -- it's 6000 / 5730.  Let's leave it as a fraction for now.

 

So use the general decay formula F = (1-r)ⁿ, where F is the fraction remaining, r is the decay rate, and n is the amount of time.  Since we're using splurgs as the unit of time, this comes out to F = (1-50%)^{(6000 / 5730)}

 

This is equal to .5 ^{(6000 / 5730)} = .4839.  Rounded to 3 decimal places, this is .484.  And that checks with our original estimate that the amount left should be a little less than half.