Carbon Dating

Carbon dating

W.F. Libby (1950)has shown that by measuring the radioactivity of ^{14}C it is possible to calculate the date at which a plant was cut or an animal died. The principle of this technique is that ^{14}C is produced in the upper atmosphere as:

^{14}_7N + ^1_0n \to ^{14}_6C + ^1_1H


There is one atom of  ^{14}C for every 7.49 \times 10^{11} carbon atom in the CO_2 of the atmosphere and in all living animals and plants. This ratio always remains constant in nature. when the animal or plant dies  ^{14}C starts disintegration as:

^{14}C \to ^{14}_7N + ^0_{-1}e \\[3mm] t = \dfrac{2.303}{\lambda} \log_{10} \dfrac{\text{Ratio of} C_{14}/ C_{12} \text{in living organism}}{\text{Ratio of} C_{14}/C_{12} \text{in dead organism}}\\[3mm] \text{or} t = \dfrac{2.303}{\lambda} \log_{10} \dfrac{\text{activity in living organism}}{\text{activity in dead organism}}


Thus by measuring the radioactivity of the substance under consideration and knowing the t_{1/2} of the ^{14}_6C  is is possible to calculate when the sample was removed from the carbon cycle.

Rock dating: use to find age of earth crust, rocks etc. it is based on the fact that U-238 decays to give Pb-206. [t_{1/2} of U-238 = 4.46 \times 10^9 years]

\dfrac{2.303}{\lambda}\log_{10}\dfrac{(U-238 + Pb-206)}{(U-238)}



\text{Radius of nucleus (r)} = (1.33 \times 10^{-15})(\text{mass number})^{1/3}m \\[3mm] \text{Density of nucleus} = \dfrac{\text{Mass No.}}{6.023 \times 10^{23}} \div \dfrac{4}{3}\pi r^2

Mass Defect (\delta m) = Expected mass – Actual mass

Binding energy, E = \Delta m. c^2 = \Delta m \times 921 MeV \\[3mm] \dfrac{-dN}{dt} = \lambda N \\[3mm] \lambda = \dfrac{2.303}{t}\log \dfrac{N_0}{N_t} \\[3mm] \lambda = \dfrac{0.693}{t_{1/2}} \\ t = 1.44 t_{1/2} = \dfrac{1}{\lambda} \\[3mm] N_t = N_0 \left(\dfrac{1}{2} \right)^n \\[3mm] \left( \text{where} n = \dfrac{\text{total time}}{\text{Half life period}} \right)

Example 1: A piece of charcoal from the ruins of a settlement in japan was found to have a ^{14}C^/{12}C ratio that was 0.617 times that found in living organisms. How old is this piece of charcoal? The half-life period of ^{14}C is 5770 years.


From eq.3,

\lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{5770 \text{yr}} = 1.201 \times 10^{-4} yr^{-1}

From eq.2,

t = \dfrac{2.303}{\lambda} \log \dfrac{N_0}{N} = \dfrac{2.303}{1.2 \times 10^{-4}} \log \dfrac{1}{0.617}

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