Question

An ancient bone contains 77g of carbon and has a decay rate of 45 counts/min. Given carbon's half-life is 5730 years, how old is it? (Assume that the ratio of isotoopes in the atmosphere is Carbon 14/Carbon 12 = 1.3*10^-12 and has remained constant.)

Answer 1

We are given
\[m = 77 g\]
\[R = 45 counts/\min=0.75counts/s\]
\[t _{1/2} = 5730 years \approx 1.81 \times 10^{11} s\]
The age of a sample is calculated using
\[\ln( R/ R _{0}) = -\lambda \times t \][1]

R0 and lambda are unknown quantities so let's find them.

\[\lambda = \ln2/t _{1/2} \][2]

lambda - disintegration constant
t1/2 - half-life in seconds NOT in YEARS. So we got it.
Next thing to calculate is R 0, which is decay rate at t = 0s or at the very beginning of existence of a sample.
\[R _{0}=\lambda \times N _{0}\][3]
We don't know N0, the number of Carbon 14 isotope atoms at the moment of time t = 0s
\[N _{0}=(initial mass-of-carbon-14)/(mass-of-1-atom-of-carbon-14)\]
[4]

Let us assume that mass of a sample through ages didn't change, cause the change in mass is negligibly small. Everything whenever created on Earth with carbon contained the carbon 14 isotope in the ratio given, thus
\[initial-mass-of-carbon-14 = 77 \times(1-1/(1+1.3 \times 10^{-12}))=1.001 \times 10^{-10}g\]
so using [4] you will get N0 then using [3] R0.
And now you are ready to find the time of life t using [1] equation.
I got 25 560 years.

Answer 2

I'm pasting again formulas that was trimmed
\[N _{0}=InitMassOfCarbon14/MassOfOneAtomOfCarbon14\]
[4]
\[InitMassOfCarbon14=77 g \times(1-1/(1+1.3 \times 10^{-12}))=1.001 \times 10^{-10} g\]

Answer 3

@Mythias