A living specimen in equilibrium with the atmosphere contains one atom of 14C (h

Question

A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5730 yr) for every 7.70 1011 stable carbon atoms. An archeological sample of wood (cellulose, C12H22O11) contains 22.8 mg of carbon. When the sample is placed inside a shielded beta counter with 88.0% counting efficiency, 844 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. ????????? yr

Explanation / Answer

We all know the formula N = No exp(-t/?), a solution of the differential equation dN/dt = -N/?. Some people prefer to write n = No exp(-kt), or other equivalent forms, but I have found the former alternative more manageable. However, all this is next to useless if we can´t determine, somehow, some crucial parameters. ? can be routinely found from N/No = exp(-t/?), or, according to given data, ½ = exp(-5 730 / ?) 2 = exp (5 730 / ?) ln 2 = 5 730 / ? ? = 5 730 / ln 2 = 8 266.64 yr. Estimation of No is more difficult: The sample contains 23.6 mg of carbon. By international agreement, one nuclidic mass u (formerly, atomic mass unit -amu) is 1/12 of the mass of one 12C atom, and equal to 1.6604 E-27 kg, which is 1.6604 E-21 mg. A 12C atom has mass =12 u, or 1.9925 E-20 mg. Thus in 23.6 mg of carbon, there are 23.6 mg / 1.9925 E-20 mg = 1.1845 E21 12C atoms. Of these, only one in every 7.70 E11 belong to the radioactive isotope 14C. So, there were No = 1.1845 E21 ÷ 7.70 E11 = 1.5383 E9 14C atoms in the sample, when the specimen still lived In order to be able to date the sample, we've yet to estimate the present number of 14C atoms in it. We can do this by solving for N in the original differential equation: N = -? dN/dt. There were 821 counts in a week. Knowing that the counter is 82.0% effective, this yields 821/0.82 = 1 001 counts per week. As there are 365/7 weeks in a year, this figure amounts to 52 206 atoms disintegrated per year. Rigurously, this rate should be negative, since the number N of 14C atoms in the sample decreases in time. Thus, N = -? dN/dt = -8 266.64 yr × -52 206 atoms/yr = 4.3157 E8 14C atoms. We can now find t from No/N = exp(t/?) t/? = ln No/N t = ? ln No/N = 8 266.64 ln (1.5383 E9 / 4.3157 E8) = 1.0507 E4 = 10 507 yr.

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