Decay constant \\\\(\\displaystyle k = \\frac {0.693}{t_{1/2}} = \\frac {0.693}{5730 \\: year} = 1.209 \ imes 10^{-4} \\: /years\\\\)The rate of counts is proportional to the number of C-14 atoms in the sample.\\\\(\\displaystyle N_0 = 100, N = 80\\\\)The age of the sample \\\\(\\displaystyle t = \\frac {2.303}{k} log\\left( \\frac {N_0}{N}\\right)\\\\)\\\\(\\displaystyle t = \\frac {2.303}{1.209 \ imes 10^{-4}} \ imes log\\left( \\frac {100}{80}\\right) = 1846 \\: years\\\\)