Half life equation model

Question

An organic sample from a Chinchorro mummy was found to contain 54.29% of the relative proportion of carbon-14 found in living matter and the atmosphere. Approximately how long before the carbon-14 measurements were made had the mummy been buried? (The half-life of carbon-14 is 5730 years.)

Answer 1

The answer provided by Arturo is certainly the conventional way to solve this problem, and is absolutely correct.  But for those of us who are somewhat "mathematically challenged", I find it easier to do it as follows:

fraction remaining = 0.5^n where n = the number of half lives that have elapsed.

Here the fraction remaining = 0.5429 and we can solve for n

0.5420 = 0.5^n

log 0.5420 = n log 0.5

-0.26529 = -0.301n

n = 0.8814 which is the number of half lives that have elapsed since burial

Since 1 half life = 5730 years...

5730 yrs x 0.8814 = 5,050 years

 

Just another way to do these kinds of problems.  I can never remember the equation A(t) = A0e-kt, so I use this.

Answer 2

This is an exponential decay problem.

 

A(t) = A₀e^-kt

 

T_1/2 = 5730 years

 

k = -ln(1/2) / T_1/2 = -ln(1/2) / 5730 years^-1 = 1.2096809 x 10^-4 years^-1

 

At current time t since burial,

 

0.5429 A₀ = A₀e^-kt

 

0.5429 = e^-kt

 

t = -ln(0.5429) / k = -ln(0.5429) / (1.2096809 x 10^-4 years^-1) ≅ 5049 years

 

It was buried about 5049 years ago.