Page 62 - Petrophysics 2E
P. 62

36     PETROPHYSICS: RESERVOIR ROCK PROPERTIES


                                                 TABLE 2.2
                        RADIOACTIVE ELEMENTS, THEIR HALF-LIVES AND  RADIOACTIVE DECAY
                                          “DAUGHTER” ELEMENTS [3]
                    Elem en  t                 Half-life               Stable Daughter

                    Carbon- 14              5,710 years                   Nitrogen-14
                    Potassium40             1.3 billion years             Argon-40
                    ThOfiUm-232             13.9 billion years            Lead-208
                    Uranium-23  5           0.7  1 billion years          Lead-207
                    Uranium-238             4.5 billion years             Lead-206



                    characterized by the following equations:


                       Ct = In (No/Nt)
                    Ctlp = In (1.0/0.5) = 0.693                                   (2.3)

                      ti12 = 0.693/C

                    where:  C  = radioactive decay constant.
                            No  = original amount of parent element.
                            Nt  = amount of daughter isotope currently present.
                            t1p = half-life of the parent element.
                            t   = age, years


                       Dating early events from the decay of carbon-14 is possible because the
                    radiocarbon is formed in the atmosphere by collision of cosmic rays with
                    nitrogen. The carbon dioxide in the atmosphere thus contains a small
                    amount  of  radiocarbon and,  therefore, all plants and animals contain
                    carbon-14 along with the stable carbon-12. When the plant or animal
                    dies, the accumulation of carbon-14 stops and its content of radiocarbon
                    decays steadily. The carbon dating is then made possible by measuring
                    the ratio of  14C to  12C in the remains of  organism and comparing it
                    to the ratio of  these isotopes in current  living plants  or animals; for
                    example,  if  the  relative  radiocarbon  content of  a specimen of  bone
                     [(14C/’2C)dead/(14C/12C)living]   is one-fourth that of the modern specimen,
                    the age of the specimen is 11,420 years. This is because 1/4 = 1/2 x 1/2
                    of two half-lives (2 half-lives x 5,710 years/half-life = 11,420 years).

                     EXAMPLE

                       If  0.35 grams of N-14 per 1.0 grams of  C-14 is found in a sediment,
                    determine the age of the sediment.
   57   58   59   60   61   62   63   64   65   66   67