How many 14 C nuclei decay per second ina 70 kgperson, assuming that the person\

Question

How many 14C nuclei decay per second ina 70 kgperson, assuming that the person's body contains 12% of carbon andthat there is one 14C nucleus for every101214C nuclei? How much time must pass after this person dies before the14C activity decreases to one decay per second? (Note we need to get a value of the half-life of14C from an outside source) How many 14C nuclei decay per second ina 70 kgperson, assuming that the person's body contains 12% of carbon andthat there is one 14C nucleus for every101214C nuclei? How much time must pass after this person dies before the14C activity decreases to one decay per second? (Note we need to get a value of the half-life of14C from an outside source)

Explanation / Answer

This problem requires many small steps. Here they are: . half life of C14 is 5730 years, which is 1.807 x1011 seconds   . This means the decay constantis .    = ln2 / T = ln2/ 1.807 x 1011 =   3.8359 x10-12 per second . the person has    12% * 70 =  8.4 kg = 8400 grams of carbon . which is    8400 grams * 1 mol per 12grams =   700 moles of carbon . Or     700 * 6.022 x1023   =   4.215 x1026 atoms of carbon   in theperson . There is only one C14 per 1012 C12, so theperson has .      4.215 x 1014 C14atoms in them . So...     rate of decay = decay constant * number of radioactive atoms = .                                   = 3.8359 x 10-12 per second * 4.215 x1014  atoms = .                          = 1617 atoms per second will decay . To answer the second question... . To answer the second question... .      rate = original rate *(1/2)t/T .        1 = 1617   (1/2)t/5730 .     ln(1/1617) = (t / 5730)ln(1/2) .        t = 5730 *ln(1/1617) / ln(1/2)   =    61100 years   must pass until the decay rate is 1 per second .        1 = 1617   (1/2)t/5730 .     ln(1/1617) = (t / 5730)ln(1/2) .        t = 5730 *ln(1/1617) / ln(1/2)   =    61100 years   must pass until the decay rate is 1 per second

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