The fossil bone has a C  14:C  12 ratio, which is 116 of that in a living animal bone. If the half-life of C  14 is 5730 years, then the age of the fossil bone is

After n half-lives (at t = nT) the number of nuclides left undecayed,N=N012nGiven, NN0=116∴116=12nor 124=12nEquating the powers, we obtainn = 4i.e., tT=4 or t=4Tor t=4 x 5730=22920 years (∵T=5730 years)