14 C is a β radioactive isotope of carbon with a half–life t½ = 5700 y. It exists in nature because — Organic Chemistry Chemistry Question
Carbon dating
14 C is a β radioactive isotope of carbon with a half–life t½ = 5700 y. It exists in nature because it is formed continuously in the atmosphere as a product of nuclear reactions between nitrogen atoms and neutrons generated by cosmic rays. We assume that the rate of formation has remained constant for thousands of years and is equal to the rate of decay, hence the amount of 14 C in the atmosphere has reached steady state. As a result 14 C accompanies the stable isotopes 12 C and 13 C in the atmosphere and participates indistinguishably in all carbon chemical reactions. It forms CO2 with oxygen and enters all living systems through photosynthesis under constant 14 C / 12 C isotopic ratio, labeling the organic molecules. This fact is used for dating samples of biological origin (e. g., silk, hair, etc.) which have been isolated by some way after the death of the organism (e.g., in an ancient grave). The 14 C / 12 C ratio in these samples does not remain constant, but decreases with time because the 14 C present is disintegrating continuously. The specific radioactivity of 14 C in living systems is 0.277 bequerel per gram of total carbon [1 Bq = 1dps (disintegration per second)]
Calculate the age of an isolated sample with a 14 C / 12 C ratio which is 0.25 that of a contemporary sample.
Model Answer
Let N0 be the 14 C/ 12 C ratio in living systems and N the same ratio found in a sample coming from a system that died t years ago. Then, the following relation between them is true: N = N0 e −λ t , where λ (= ln 2 / t½) the disintegration constant for 14 C. The above equation becomes
ln(N / No) = -λ t ⇒ ln 0.25 = - (ln 2 / 5700 y) t
t = 11400 y
What happens to a 14 C atom when it disintegrates?
Model Answer
The β decay scheme is based on the nuclear reaction n = p + – + ve– where p is a proton and ve– an electron antineutrino. In the case of 14 C we have
14 C → 14 N + β – + ve–
hence C becomes a (common) 14 N atom.
What do you expect will happen to a 14 C containing organic molecule (e.g., DNA, protein, etc.) of a living organism when this 14 C atom disintegrates.
Model Answer
If an organic molecule contains 14 C, the consequence of its disintegration can be grave for the structure of the molecule, causing great damage to the molecule in the vicinity of the 14 C atom. At least the chemical bond is raptured since 14 N is a chemically different atom than 14 C. Free radicals may also be created.
Calculate the radioactivity of a 75 kg human body due to 14C and the number of 14 C atoms in the body, given that the amount of total carbon is about 18.5 %.
Model Answer
The total carbon inside a human body of 75 kg is 75 kg x 0.185 = 13.9 kg. The total radioactivity (R) is R = 0.277 Bq g –1 × 13.9 kg = 3850 Bq. The amount of 14 C present is estimated from the total radioactivity as follows:
R = – d N / d t = λ N
Then
N = R / λ = (3850 s^-1 × 5700 y × 365.25 d × 24 h × 60 min × 60 s) / ln 2 = 1.00 × 10^15 atoms = 1.66 nmol