Q10(a). Radiocarbon age of a piece of wood
The amount of C-14 isotope in a piece of wood is found to be one-sixth of its amount present in a fresh piece of wood. Calculate the age of the wood. Given that half-life of C-14 =5577 years.
Answer in brief
Age =1.44×10⁴ years (14 416 years).
Key points
- The assumption is that the ratio ¹⁴C/¹²C in the atmosphere has been constant; the result is in radiocarbon years, normally calibrated with tree-ring and coral data.
- The count of one-sixth (16.7%) is within the limit of the method (about 50 000 y).
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
Radioactive decay is first order: N=N₀e^(-λ t), with λ=0.693t_(1/2)=(0.693)/(5577)=1.243×10⁻⁴ y⁻¹.
t=(1)/(λ)ln(N₀)/(N)=2.3031.243×10⁻⁴log 6=2.303×0.77821.243×10⁻⁴=1.44×10⁴ years
Check by half-lives: (N)/(N₀)=(12)^t/t_(1/2), tt_(1/2)=₂ 6=2.585, so t=2.585×5577=14 416 y: between 2 half-lives (4 → 1/4) and 3 half-lives (1/8).
- The assumption is that the ratio ¹⁴C/¹²C in the atmosphere has been constant; the result is in radiocarbon years, normally calibrated with tree-ring and coral data.
- The count of one-sixth (16.7%) is within the limit of the method (about 50 000 y).
wood stops exchanging carbon with the atmosphere at its death; ¹⁴C (β⁻, 0.156 MeV) then decays, so counting it gives the time since death.
Coversradiocarbon datingRadiocarbon age of a piece of wood
All parts of Question 10
- 10(a) Radiocarbon age of a piece of wood10 marks
- 10(b) Identify M from the CO stretching frequencies10 marks
- 10(c) Why Al cannot be titrated directly with EDTA10 marks