Question

A living specimen in equilibrium with the atmosphere contains one atom of ¹⁴C (half-life = 5730 yr) for every 7.7 ✕ 10¹¹ stable carbon atoms. An archeological sample of wood (cellulose, C₁₂H₂₂O₁₁) contains 20.6 mg of carbon. When the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 765 counts are accumulated in one week. Assuming that the cosmic-ray flux and the Earth's atmosphere have not changed appreciably since the sample was formed, find the age of the sample.

Answer 1

Total number of carbon atoms in given sample

N = 20.6e-3 g * 6.02e23 molecules / mol / 12 g/mol

N = 1.033e21 atoms

one in these atoms is C-14 so,

Number of C-14

N' = 1.033e21 / 7.7e11

N' = 1.342e9

Now, we find decay constant

= ln 2 / T_1/2

= ln 2 / 5730

= 3.83e-12 / sec

Initial activity

R = 3.83e-12 * 1.342e9 * 7 * 86400

R = 3108.587 decays / week

at some given time

R' = 765 / 0.88 = 869.32 decays / week

Now, from decay equation

ln ( R' / R ) = - t

so,

t = -1/ * ln ( R' / R)

t = -1/ 3.83e-12 * ln ( 869.32 / 3108.587)

t = 3.326e11 seconds

or

t = 1.053e4 years