A sample of a human bone found in a cave has a 14C/12C ratio that is 12% of the ratio found in living humans. How old is the bone in years? The half-life of 14C = 5770 years. (a) 2.1 x 104 (b) 4.0 x 105 (c) 1.0 x 105 (d) 1.8 x 104 (e) 4.3 x 10-4 Solution Given: Half life = 5770 years use relation between rate constant and half life of 1st order reaction k = (ln 2) / k = 0.693/(half life) = 0.693/(5770) = 1.201*10^-4 years-1 we have: [14C]o = 100.0 (Let initial ratio be 100) [14C] = 12.0 k = 1.201*10^-4 years-1 use integrated rate law for 1st order reaction ln[14C] = ln[14C]o - k*t ln(12) = ln(100) - 1.201*10^-4*t 2.4849 = 4.6052 - 1.201*10^-4*t 1.201*10^-4*t = 2.1203 t = 1.8*10^4 years Answer: 1.8*10^4 years .