Page 188 - Essentials of physical chemistry
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150                                                  Essentials of Physical Chemistry


                                               Po       Bi       Pb decay
                             1200

                             1000
                             800
                            Atoms  600


                             400

                             200

                               0
                                 0     10     20     30     40     50     60
                                                    Hours
            FIGURE 7.6  A plot of the successive decay of  204 Po !  204 Bi with t 1=2 ¼ 3:53 h followed by  204  Bi !  204 Pb
                                               84
                                                      83
                                                                                83
                                                                                       82
            with t 1=2 ¼ 11:2 h based on an initial 1000 atoms of  204 Po. Thanks to Prof. Steven Yates of the University of
                                                  84
            Kentucky for suggesting this example.
            but if mole quantities are involved in a chemical reaction we might have to use more significant
            figures in the half-life values than just 0.693. Even so we have solved the problem and you can
            judge whether more or less significant figures are warranted for a particular case. Note that we have
            solved the numerical example for the case where k a > k b and the temporary maximum in tank B will
            occur under that situation. If k a < k b , the temporary buildup in tank B will not occur.

              A ! B ! C Water Tank Summary:


                 N 0 ¼ 1000 gal;  t 1=2 (A) ¼ 3 days;  t 1=2 (B) ¼ 8 days;  t max N B  ¼ 6:7936 days
                                      ;                       ;
                 N A ¼ 208:19 gal at t max N B  N B ¼ 555:16 gal at t max N B  N C ¼ 236:65 gal at t max N B

            Now that we have some understanding of the way in which the decay scheme works, we can look
            back and understand some of the difficulty Marie Curie had in isolating radioactive polonium
            (named for her home country of Poland). Using modern data from Ref. [1], we find t 1=2 ¼ 3:53 h for
            the electron capture of  204  204 Bi followed by another electron capture by  204  204 Pb with
                              84  Po !  83                                 83  Bi !  82
            t 1=2 ¼ 11:2 h to form stable  204 Pb. Although other isotopes are involved, this scheme shows the
                                   82
            difficulty in isolating Po from Bi while the decay process is going on. Note that the time scale in
            Figure 7.6 is in hours.
              Considering the many applications of this type of problem such as nuclear decay and various
            forms of time-dependent spectroscopy (NMR, UV–VIS, etc.) there is sufficient detail to the solution
            presented above to allow it to be used in a number of situations and it is certainly one of the
            ‘‘essential’’ aspects of basic kinetics in physical chemistry.

            SPLITTING THE ATOM

            After the work by Marie Curie and her daughter Irene established a new field of research in the
            radioactivity of elements, others carried out similar experiments to begin to understand the internal
            structure of the nucleus and a period of increased research occurred in the 1930s. At the Kaiser
            Wilhelm Institute in Berlin and the Niels Bohr Institute in Stockholm, a drama unfolded in 1938 that
            ushered in the atomic age. Lise Meitner was a petite, shy Austrian girl who made friends with a
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