Unraveling the Mystery of Carbon Dating: Age of an Ancient Fire Pit

[Jon1527]

Carbon dating!

A 5g charcoal sample from an ancient fire pit has a 14C activity of 63 disintegrations
per minute. A living tree has a 14C activity of 15 disintegrations per minute per 1g.
The half-life of 14C is 5730 years. How old is the charcoal sample from the ancient fire
pit?

Homework Equations

dont think that any are necessary (I think!)

The Attempt at a Solution

63/5 =12.6
12.6/15 = 0.84
5730*0.84 = 4813.2 years

This isn't the right answer I am sure. This is taken from a previous exam paper I am doing to revise for my exams, and carbon dating is defintly gona come up, but can't find a method for doing this kind of question any where can somone pleeease help!

 

[mgb_phys]

You haven't understood half-life correctly.
The activity is halved every 5730years, so if you start with 15/s then after 5370 years you will have 7.5/s and after 10740 years 3.75/s.
If you draw this on a graph you will see that it isn't a stright line.

If you haven't studied enough maths to work this out the exam will normally 'cheat' and use answers that are whole numbers of half-lives. Because this doesn't you should have come across the equation for exponential decay.
Look up half-life or exponential decay.

 

[andrevdh]

When the tree is alive it absorbs carbon (dioxide) from the atmosphere. This keeps the ratio of radioactive carbon in it (per gram) constant. When it dies the absorption (respiration) process stops and the remaining radioactive carbon (14) starts to decrease due to decay.

The given data gives you the half-life of the decay process. The relation between the half-life and the decay constant is

$$T_{1/2}\ \lambda = \ln(2)$$

so the initial activity is 15 disintegrations per minute per gram. The question requires you calculate the amount of time that elapsed to bring it down to 12.6. The decay decreases exponentially with time.

 

[Chronos]

Agreed. The formula for radioactive decay was logarithmic last time I checked.

 

[andrevdh]

Yes, the exponential formula can be changed into a (natural) logarithmic one that is linear in time.

 

[mgb_phys]

I assumed that since the OP wasn't given simple times they must have studied decay laws and was trying to give hints on what to look up.
If they haven't studied decay laws then quoting formulae with log(2) and lamba weren't going to help.

 

[andrevdh]

I am not sure how one would do this without decal laws. What other approach is there? Using half-lifes?

$$\frac{A_o}{2^n} = A_{now}$$

 

[mgb_phys]

If this is in an intro course before they have studied the necessary maths to use log funtions the decay rate is often chosen to be a whole number of half-lives, or you draw a graph and pick numbers off the curve.

 

[andrevdh]

To my knowledge the decay rate (or activity) is

$$A = \frac{dN}{dt} = -\lambda\ N$$

which also decreases exponentially with time.

 

[trex55]

this is how the solution should be,
Equation : R = R₀e^-λt

t = (1/λ) (ln R₀/R) = (5730 y/ ln 2) ln[(15.3/63.0)(5.00/1.00)] = 1.61x10³ y