Question

Carbon-14 has a half life of 5730 years. If a shell is found and has 52% of its original Carbon-14, how old is it?

Answer 1

Approximately `5406` years

Explanation:

`-5730 * log_2(52/100) = -5730 * log(52/100)/log(2) ~~ 5406` years

The proportion of `""^14C` remaining, compared with the original is modelled by the function:

`p(t) = 2^-t/(5730)`

where `t` is measured in years.

Given that `p(t) = 52/100`, we have:

`52/100 = 2^(-t/5730)`

Taking logs base `2` we have:

`log_2(52/100) = -t/5730`

So multiplying both sides by `-5730` and transposing we get the formula:

`t = -5730*log_2(52/100)`

Footnote

The complexity of radiocarbon dating comes from two factors:

• The proportion of `""^14C` to `""^12C` is very small to start with.
• The proportion has been dramatically affected by factors like the industrial revolution, which replaced a significant portion (one third? one quarter?) of the `CO_2` in the atmosphere with carbon dioxide generated from fossil fuels - which contain virtually no remaining `""^14C`. As a result the method needs calibration to provide accurate dates.

Answer 2

`sf(5409color(white)(x)"yr")`

Explanation:

The expression for radioactive decay is:

`sf(N_t=N_0e^(-lambdat))`

`sf(N_0)` is the initial number of undecayed atoms

`sf(N_t)` is the number of undecayed atoms at time `sf(t)`

`sf(lambda)` is the decay constant

Taking natural logs of both sides:

`sf(lnN_t=lnN_0-lambdat)`

`:.` `sf(ln(N_t/N_0)=-lambdat)`

We can get the value of `sf(lambda)` using the expression:

`sf(lambda=0.693/t_(1/2)=0.693/5730=1.209xx10^(-4)color(white)(x)a^(-1))`

Putting in the numbers:

`sf(ln(52/100)=-1.208xx10^(-4)xxt=-0.6539)`

`:.` `sf(t=0.6539/(1.209xx10^(-4))=5409color(white)(x)yr)`