SlideShare une entreprise Scribd logo
1  sur  13
Chapter 39. Nuclear Physics and the Nucleus                                              Physics, 6th Edition


                 Chapter 39. Nuclear Physics and the Nucleus

The Elements
                                                     208
39-1. How many neutrons are in the nucleus of        82    Pb ? How many protons? What is the ratio

       N/Z?     (N is the number of neutrons and Z is the number of protons.)

                                                                                        A
            A = N + Z;      N = A – Z = 126 neutrons;          Z = 82 protons;            = 1.54
                                                                                        Z


39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol

       for this nucleus.

                                                                          235
                           A = N + Z = 143 + 92 = 235; Z = 92:            92    U


39-3. From a stability curve it is determined that the ratio of neutrons to protons for a cesium

       nucleus is 1.49. What is the mass number for this isotope of cesium?

                            N
              Z = 55;         = 1.49;   N = 1.49(55) = 81.95; A = N + Z ;           A = 137
                            Z


39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by
                                            1
                                   r = r0 A 3    r0 = 1.2 x 10−15 m

                                                                           197
       What is the approximate radius of the nucleus of a gold atom        79    Au ?

                   r = (1.2 x10−15 m) 3 197 = 6.98 x 10-15 m ;     r = 6.98 x 10-15 m


39-5. Study Table 39-4 for information on the several nuclides. Determine the ratio of N/Z for

       the following nuclides: Beryllium-9, Copper-64, and Radium 224.

                                                                         N
                        Beryllium: A = 9; Z = 4; N = 9 – 4 = 5;            = 1.25
                                                                         Z



                                                  26
Chapter 39. Nuclear Physics and the Nucleus                                      Physics, 6th Edition


                                                                          N
39-5. (Cont.)         Copper: A = 64; Z = 29; N = 64 – 29 = 35;             = 1.21
                                                                          Z

                                                                              N
                      Radium: A = 224; Z = 88; N = 224 – 88 = 136;              = 1.55
                                                                              Z


The Atomic Mass Unit

39-6. Find the mass in grams of a gold particle containing two million atomic mass units?

                                    1.66 x 10-27 kg 
                     m = 2 x 106 u                  ;   m = 3.32 x 10-21 kg
                                        1.00 u      

39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules?

                                       1u       
                       m = 2 kg             -27 ;       m = 1.20 x 1027 u
                                 1.6606 x 10 kg 

                                       931 MeV 
                   m = 1.204 x 1027 u          ;        m = 1.12 x 1030 MeV
                                       1u 

                                  30  1.6 x 10-13 J 
                   m = 1.12 x 10 MeV                ;      m = 1.79 x 1017 J
                                      1 MeV 


39-8. A certain nuclear reaction releases an energy of 5.5 MeV. How much mass (in atomic

       mass units) is required to produce this energy?

                                       1u      
                          m = 5.5 MeV          ;        m = 0.00591 u
                                       931 MeV 


39-9. The periodic table gives the average mass of a silver atom as 107.842 u. What is the

       average mass of the silver nucleus?    (Recall that me = 0.00055 u )

       For silver, Z = 47. Thus, the nuclear mass is reduced by the mass of 47 electrons.

                      m = 107.842 u – 47(0.00055 u);         m = 107.816 u



                                                27
Chapter 39. Nuclear Physics and the Nucleus                                        Physics, 6th Edition


*39-10. Consider the mass spectrometer as illustrated by Fig. 39-3. A uniform magnetic field of

        0.6 T is placed across both upper and lower sections of the spectrometer., and the electric

        field in the velocity selector is 120 V/m. A singly charged neon atom (+1.6 x 10-19 C) of

        mass 19.992 u passes through the velocity selector and into the spectrometer. What is the

        velocity of the neon atom as it emerges from the velocity selector?

                 E 120, 000 V/m
            v=     =            ;       v = 2 x 105 m/s
                 B     0.6 T

                                                                               R
*39-11. What is the radius of the circular path followed by the

          neon atom of Problem 19-10?
                                                                            B = 0.6 T into paper
                        1.66 x 10-27 kg            −26
          m = 19.992 u                   = 3.32 x10 kg
                             1u         

          mv 2              mv           (3.32 x 10-27 kg)(2 x 105 m/s)
               = qvB;    R=    ;      R=                                ;   R = 6.92 cm
           R                qB              (1.6 x 10-19 C)(0.600 T)


Mass Defects and Binding Energy
(Refer to Table 39-4 for Nuclidic Masses.)

                                                                              20
*39-12. Calculate the mass defect and binding energy for the neon-20 atom     10   Ne .

           mD = [( ZmH + Nmn )] − M = [ 10(1.007825 u) + 10(1.008665 u)] − 19.99244 u

                        mD = 20.016490 u –19.99244 u ;      mD = 0.17246 u

                                        931 MeV 
              E = mD c 2 = (0.17246 u)           = 160.6 MeV ;       E = 161 MeV
                                        1u      

                                                                                    3
*39-13. Calculate the binding energy and the binding energy per nucleon for tritium 1 H . How

        much energy in joules is required to tear the nucleus apart into its constituent nucleons?

            mD = [( ZmH + Nmn )] − M = [ 1(1.007825 u) + 2(1.008665 u) ] − 3.016049 u



                                                 28
Chapter 39. Nuclear Physics and the Nucleus                                             Physics, 6th Edition


                                                                931 MeV 
39-13. (Cont.) mD = 0.009106 u;      E = mD c 2 = (0.009106 u)           = 8.48 MeV
                                                                1u      

                                   EB 8.48 MeV            EB
               E = 8.48 MeV;         =         ;             = 2.83 MeV/nucleon
                                   A      3               A

            Energy to tear apart: EB = 8.48 MeV(1.6 x 10-13 J/MeV) = 1.36 x 10-12 J


*39-14. Calculate the mass defect of 7 Li . What is the binding energy per nucleon?
                                     3



            mD = [( ZmH + Nmn )] − M = [ 3(1.007825 u) + 4(1.008665 u) ] − 7.016930 u

                       mD = 7.058135 u –7.016930 u ;      mD = 0.041205 u

                                         931 MeV 
              E = mD c 2 = (0.041205 u)           = 38.4 MeV ;        E = 38.4 MeV
                                         1u      

                          EB 38.4 MeV             EB
                            =         ;              = 5.48 MeV/nucleon
                          A      7                A


                                                                   12
*39-15. Determine the binding energy per nucleon for carbon-12 (   6    C ).

                  mD = [ 6(1.007825 u) + 6(1.008665 u) ] − 12.0000 u = 0.09894 u

                                    931 MeV       EB 92.1 MeV
          E = mD c 2 = (0.09894 u)          ;       =         = 7.68 MeV/nucleon
                                    1u            A      12


                                                                               197
*39-16. What is the mass defect and the binding energy for a gold atom         79    Au ?

              mD = [ 79(1.007825 u) + 118(1.008665 u) ] − 196.966541 u = 1.674104 u

                                                  931 MeV 
                       E = mD c 2 = (1.674104 u)           = 1.56 GeV;
                                                  1u      

                              EB 1.56 GeV
                                =         = 7.91 MeV/nucleon
                              A     197




                                                  29
Chapter 39. Nuclear Physics and the Nucleus                                                        Physics, 6th Edition

                                                                               120
*39-17. Determine the binding energy per nucleon for tin-120 (                 50    Sn ).

              mD = [ 50(1.007825 u) + 70(1.008665 u) ] − 119.902108 u = 1.09569 u

                  EB  (1.09569 u)(931 MeV/u                         EB
                    =                       ;                          = 8.50 MeV/nucleon
                  A            120                                  A


Radioactivity and Nuclear Decay

39-18. The activity of a certain sample is rated as 2.8 Ci. How many nuclei will have

        disintegrated in a time of one minute?

                                 3.7 x 1010s-1 
                Nuclei = 2.8 Ci                 (60 s);               nuclei = 6.22 x 1012 nuclei
                                     1 Ci      


                                   60
39-19. The cobalt nucleus          27   Co emits gamma rays of approximately 1.2 MeV. How much mass

        is lost by the nucleus when it emits a gamma ray of this energy?

                                                1u      
                                   E = 1.2 MeV          ;              m = 0.00129 u
                                                931 MeV 


39-20. The half-life of the radioactive isotope indium-109 is 4.30 h. If the activity of a sample is

        1 mCi at the start, how much activity remains after 4.30, 8.60, and 12.9 h?

                             t / T½                                                  1
                     1                     t   4.3 h                         1
              R = R0                 ;       =       = 1;        R = (1 mCi)   ;         R = 0.5 mCi
                     2                    T½ 4.3 h                           2

                          t / T½                                                  2
                    1                     t   8.6 h                      1
             R = R0              ;          =       = 2;     R = (1 mCi)   ;             R = 0.25 mCi
                    2                    T½ 4.3 h                        2

                          t / T½                                                     3
                    1                     t 12.9 h                           1
             R = R0              ;         =       = 3;          R = (1 mCi)   ;         R = 0.125 mCi
                    2                    T½ 4.3 h                            2




                                                              30
Chapter 39. Nuclear Physics and the Nucleus                                                       Physics, 6th Edition


39-21. The initial activity of a sample containing 7.7 x 1011 bismuth-212 nuclei is 4.0 mCi. The

        half-life of this isotope is 60 min. How many bismuth-212 nuclei remain after 30 min?

        What is the activity at the end of that time?

                       t / T½                                                           ½
                 1               t   30 min                          1
          N = N0              ;    =        = 0.5; N = (7.7 x 1011 )   = 5.44 x 1011 nucl.
                 2              T½ 60 min                            2

                                         t / T½                            ½
                              1                                 1
                       R = R0                   ;   R = (4 mCi)   ;           R = 2.83 mCi
                              2                                 2


*39-22. Strontium-90 is produced in appreciable quantities in the atmosphere during a nuclear

        explosion. If this isotope has a half-life of 28 years, how long will it take for the initial

        activity to drop to one-fourth of its original activity?

                                                       n               n
                                             1              R 1    1
                                      R = R0   ;              =  = ; n=2
                                             2              R0  2  4

                                       t   t
                                 n=      =     = 2;             t = 2(28 yr); t = 56 yr
                                      T½ 28 yr


*39-23. Consider a pure, 4.0-g sample of radioactive Gallium-67. If the half-life is 78 h, how

        much time is required for 2.8 g of this sample to decay?

                  When 2.8 g decay, that leaves 4 g – 2.8 g or 1.20 g remaining.

                                           n                                        n
                                1                   m 1.2 g                    1
                         m = m0   ;                    =    = 0.300;             = 0.300
                                2                   m0   4g                    2

              The solution is accomplished by taking the common log of both sides:

                                                                                        −0.523
                n log(0.5) = log(0.300);                   − 0.301n = −0.523;     n=           = 1.74
                                                                                        −0.301

                                  t   t
                         n=         =    = 1.74;               t = 1.74 (78 h)      t = 135 h
                                 T½ 78 h



                                                                31
Chapter 39. Nuclear Physics and the Nucleus                                                        Physics, 6th Edition


*39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life?

                      n
             1 1
              =  ;           0.200 = (0.5) n ; n log(0.5) = log(0.2);                − 0.301 n = −0.699
             5 2

                                 t 10 h                                10 h
                          n=       =    = 2.32;               T½ =          ;       T½ = 4.31 h
                                T½   T½                                2.32


Nuclear Reactions

*39-25. Determine the minimum energy released in the nuclear reaction

                                             19
                                             9    F +1 H →4 He + 16 O + energy
                                                     1
                                                          2      8



        The atomic mass of         19
                                   9    F is 18.998403 u,          4
                                                                   2   He = 4.002603 u,     1
                                                                                            1H = 1.007824 u.

                   E=     19
                          9    F + 1H - 2 He - 16 O ; (Energy comes from mass defect)
                                   1    4
                                               8



               E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u

                                                931 MeV 
                               E = (0.00871 u)          ;                     E = 8.11 MeV
                                                1u      


*39-26. Determine the approximate kinetic energy imparted to the alpha particle when

        Radium-226 decays to form Radon-222. Neglect the energy imparted to the radon

        nucleus.

                      226
                      88    Ra →        222
                                        86    Rn + 4 He + Energy;
                                                   2
                                                                            226
                                                                            88    Ra = 226.02536

                                                                931 MeV 
       E = 226.02536 u - 222.017531 u - 4.002603 u = 0.00523 u           = 4.87 MeV
                                                                1u      


*39-27. Find the energy involved in the production of two alpha particles in the reaction

                                         7
                                         3   Li + 1H → 2 He + 2 He + energy
                                                  1    4      4




                                                              32
Chapter 39. Nuclear Physics and the Nucleus                                                   Physics, 6th Edition


                                                                 931 MeV 
       E = 7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u           = 17.3 MeV
                                                                 1u      

*39-28. Compute the kinetic energy released in the beta minus decay of thorium-233.

          233
          90    Th →   233
                       91    Pa + 0 β + energy;
                                  +1
                                                     233
                                                     90    Th = 233.041469 u;    233
                                                                                 91    Pa = 233.040130

                       E = 233.041469 u - 233.040130 u - 0.00055 u = 0.000789 u

                                               931 MeV 
                               E = 0.000789 u          ;           E = 0.735 MeV
                                               1u      


*39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus

                       17
        producing      8    O and 1H? (17 O = 16.999130 u) .
                                  1    8


                                        4
                                        2   He + 14 N + Energy →
                                                 7
                                                                   17
                                                                   8
                                                                            1
                                                                        O + 1H

                                                                            931 MeV 
        E = 16.999130 u + 1.007825 u - 14.003074 - 4.002603 u = 0.001278 u          
                                                                            1u      

                  E = 1.19 MeV               This is the Threshold Energy for the reaction



Challenge Problems

*39-30. What is the average mass in kilograms of the nucleus of a boron-11 atom?

                                             1.66 x 10-27 kg 
                            m = 11.009305 u                  ;        m = 1.83 x 10-26 kg
                                                 1.00 u      


*39-31. What are the mass defect and the binding energy per nucleon for boron-11?

            mD = [( ZmH + Nmn )] − M = [ 5(1.007825 u) + 6(1.008665 u) ] − 11.009305 u

                             mD = 11.09112 u –11.009305 u ;             mD = 0.08181 u

                                           931 MeV 
                 E = mD c 2 = (0.08181 u)           = 76.2 MeV ;               E = 76.2 MeV
                                           1u      


                                                            33
Chapter 39. Nuclear Physics and the Nucleus                                            Physics, 6th Edition


                            EB 76.2 MeV            EB
                              =         ;             = 6.92 MeV/nucleon
                            A      11              A

*39-32. Find the binding energy per nucleon for Thallium-206.

              mD = [ 81(1.007825 u) + 125(1.008665 u) ] − 205.976104 u = 1.740846 u

                                    931 MeV           EB 1621 MeV
         E = mD c 2 = (1.740846 u)          ;           =         = 7.87 MeV/nucleon
                                    1u                A     206



*39-33. Calculate the energy required to separate the nucleons in mercury-204.

               mD = [ 80(1.007825 u) + 124(1.008665 u) ] − 203.973865 u = 1.7266 u

                                               931 MeV 
                      E = mD c 2 = (1.7266 u)          ;     EB = 1610 MeV
                                               1u      



*39-34. The half-life of a radioactive sample is 6.8 h. How much time passes before the activity

        drops to one-fifth of its initial value?

                        n
          R 1 1
            = =  ;         0.200 = (0.5) n ; n log(0.5) = log(0.2);         − 0.301 n = −0.699
          R0 5  2 

                 0.699              t   6.8 h                       6.8 h
            n=         = 2.32; n =    =       = 2.32;        T½ =         ;       T½ = 2.93 h
                 0.301             T½    T½                         2.32




*39-35. How much energy is required to tear apart a deuterium atom? (1 H = 2.014102 u)
                                                                     2




                 mD = [ 1(1.007825 u) + 1(1.008665 u) ] − 2.014102 u = 0.002388 u

                                               931 MeV 
                    E = mD c 2 = (0.002388 u)          ;          EB = 2.22 MeV
                                               1u      




                                                   34
Chapter 39. Nuclear Physics and the Nucleus                                                           Physics, 6th Edition




*39-36. Plutonium-232 decays by alpha emission with a half-life of 30 min. How much of this

        substance remains after 4 h if the original sample had a mass of 4.0 g? Write the equation

        for the decay?

                           t / T½                                                           8
                     1                     t   4h                                 1
              m = m0              ;          =     = 8.00;              m = (4 g)   ;       m = 15.6 mg
                     2                    T½ 0.5 h                                2

                                                232
                                                94    Pu →   228
                                                             92    U + 4 He + energy
                                                                       2




*39-37. If 32 x 109 atoms of a radioactive isotope are reduced to only 2 x 109 atoms in a time of

        48 h, what is the half-life of this material?

                                        n
        N    2 x 109  1 
           =        9
                      =   ; 0.0625 = (0.5) n ; n log(0.5) = log(0.0625); − 0.301 n = −1.204
        N 0 32 x 10  2 

                  1.204              t   48 h                                        48 h
             n=         = 4.00; n =    =      = 4.00;                         T½ =        ;      T½ = 12.0 h
                  0.301             T½    T½                                         4.00


*39-38. A certain radioactive isotope retains only 10 percent of its original activity after a time of

        4 h. What is the half-life?

                           n
           N    1 1
              =  =   ; 0.100 = (0.5) n ; n log(0.5) = log(0.100); − 0.301 n = −1.00
           N 0 10  2 

                  1.00               t   4h                                        4h
             n=         = 3.32; n =    =    = 3.32;                         T½ =        ;       T½ = 72.3 min
                  0.301             T½ T½                                          3.32

*39-39. When a 6 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are
               3


        released. Write the equation for this reaction. What is the net energy transfer in this case?

                                            6
                                            3   Li + 1H → 2 He + 3 He + energy
                                                     1    4
                                                                 2




                                                                     35
Chapter 39. Nuclear Physics and the Nucleus                                                     Physics, 6th Edition


                                                                      931 MeV 
     E = 6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u           = 4.02 MeV
                                                                      1u      

*39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate

        the disintegration energy.

                                      238
                                      92    U→   224
                                                 90
                                                            4
                                                       Th + 2 He + energy

                     E = 238.05079 u - 234.04363 u - 4.002603 u = 0.00456 u

                                          931 MeV 
                           E = 0.00456 u          ;                  E = 4.24 MeV
                                          1u      


*39-41. A 9-g sample of radioactive material has an initial activity of 5.0 Ci. Forty minutes later,

        the activity is only 3.0 Ci. What is the half-life? How much of the pure sample remains?

                           n
         R 3 Ci  1 
           =    =  ;          0.600 = (0.5) n ; n log(0.5) = log(0.6);                 − 0.301 n = −0.222
         R0 5 Ci  2 

             0.222               t   40 min                                   40 min
        n=         = 0.737; n =    =        = 0.737;                   T½ =          ;       T½ = 54.3 min
             0.301              T½     T½                                      0.737

                                             n                 0.737
                                      1         1
                               m = m0   = (9 g)                      m = 5.40 g
                                      2         2



Critical Thinking Problems

*39-42. Nuclear fusion is a process that can produce enormous energy without the harmful

        byproducts of nuclear fission. Calculate the energy released in the following nuclear

        fusion reaction:

                                  3
                                  2   H + 3 He →2 He +1 H+1H + energy
                                          2
                                                4     1   1




                                                                          931 MeV 
             E = 2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u          
                                                                          1u      


                                                          36
Chapter 39. Nuclear Physics and the Nucleus                                    Physics, 6th Edition


                                         E = 12.9 MeV



*39-43. Carbon-14 decays very slowly with a half-life of 5740 years. Carbon dating can be

       accomplished by seeing what fraction of carbon-14 remains, assuming the decay process

       began with the death of a living organism. What would be the age of a chunk of charcoal

       if it was determined that the radioactive C-14 remaining was only 40 percent of what

       would be expected in a living organism?

                        n
        R           1
           = 0.40 =   ;    0.400 = (0.5) n ; n log(0.5) = log(0.4);   − 0.301 n = −0.399
        R0          2

             0.399               t    t
        n=         = 0.737; n =    =       = 1.32;        t = 1.32(5740 yr); t = 7590 yr
             0.301              T½ 5740 yr


*39-44. The velocity selector in a mass spectrometer has a magnetic field of 0.2 T perpendicular

       to an electric field of 50 kV/m. The same magnetic field is across the lower region. What

       is the velocity of singly charged Lithium-7 atoms as they leave the selector? If the radius

       of the path in the spectrometer is 9.10 cm, find the atomic mass of the lithium atom?

                  E 50, 000 V/m
             v=     =           ;     v = 2.5 x 105 m/s
                  B     0.2 T

         mv 2            qBR      (1.6 x 10-19 C)(0.2 T)(0.091 m)                R
              = qvB; m =     ; m=                                 ;
          R               v               (2.5 x 105 m/s)

                                        1u      
             m = 1.165 x 10-26 kg           -27 ;   m = 7.014 u            B = 0.2 T into paper
                                   1.66 x 10 kg 


*39-45. A nuclear reactor operates at a power level of 2.0 MW. Assuming that approximately

       200 MeV of energy is released for a single fission of U-235, how many fission processes

       are occurring each second in the reactor?


                                                37
Chapter 39. Nuclear Physics and the Nucleus                                                 Physics, 6th Edition


                 2 x 106 J/s                        1.25 x 1019 MeV/s
        P=                     = 1.25 x 1019 MeV/s;                   = 6.25 x 10216 fissions/s
             1.6 x 10-19 J/MeV                      200 MeV/fission

                                                         14
*39-46. Consider an experiment which bombards            7    N with an alpha particle. One of the two


        product nuclides is 1 H . The reaction is
                            1


                                        4
                                        2   He +14 N → Z X +1 H
                                                7
                                                       A
                                                            1


        What is the product nuclide indicated by the symbol X? How much kinetic energy must

        the alpha particle have in order to produce the reaction?

                Conservation of nucleons means:
                                                          4
                                                          2   He +14 N + enery →
                                                                  7
                                                                                     17
                                                                                     8    O +1 H
                                                                                             1




              E = 16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 u

                                          931 MeV 
                          E = 0.001282 u          ;              E = 1.19 MeV
                                          1u      


*39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius

        of the path followed by 7 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 6 Li
                                3                                                            3


                                                                            6
        (6.0151 u). What is the radius of the path followed by the          3   Li isotopes?

                    mv 2            mv
                         = qvB; R =    ; (See Problems 39-10 and 39-11)
                     R              qB

                  m1 R1               m2 R1 (6.0151 u)(14 cm)
                    = ;        R2 =        =                  ;            R2 = 6.92 cm
                  m2 R2                m1       7.0169 u




                                                    38

Contenu connexe

Tendances

Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...
Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...
Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...IOSR Journals
 
Wave Optics Class 12 Part-1
Wave Optics Class 12 Part-1Wave Optics Class 12 Part-1
Wave Optics Class 12 Part-1Self-employed
 
Standing waves
Standing wavesStanding waves
Standing wavesJenny He
 
nuclear binding energy
 nuclear binding energy nuclear binding energy
nuclear binding energyZeeshan Khalid
 
Ch 31 Nuclear Physics and Radioactivity
Ch 31 Nuclear Physics and RadioactivityCh 31 Nuclear Physics and Radioactivity
Ch 31 Nuclear Physics and RadioactivityScott Thomas
 
General Properties of Nuclear
General Properties of NuclearGeneral Properties of Nuclear
General Properties of NuclearUsydntprtty
 
Chapter 6.1 : Introduction to Chemical Bonding
Chapter 6.1 : Introduction to Chemical BondingChapter 6.1 : Introduction to Chemical Bonding
Chapter 6.1 : Introduction to Chemical BondingChris Foltz
 
Chapter 21 Lecture- Nuclear Chemistry
Chapter 21 Lecture- Nuclear ChemistryChapter 21 Lecture- Nuclear Chemistry
Chapter 21 Lecture- Nuclear ChemistryMary Beth Smith
 
IB Chemistry Limiting, Excess, Theoretical and Percentage Yield
IB Chemistry Limiting, Excess, Theoretical and Percentage YieldIB Chemistry Limiting, Excess, Theoretical and Percentage Yield
IB Chemistry Limiting, Excess, Theoretical and Percentage YieldLawrence kok
 
Class XII Solutions
Class XII   Solutions Class XII   Solutions
Class XII Solutions Arunesh Gupta
 
Strongly Interacting Atoms in Optical Lattices
Strongly Interacting Atoms in Optical LatticesStrongly Interacting Atoms in Optical Lattices
Strongly Interacting Atoms in Optical LatticesABDERRAHMANE REGGAD
 
Introduction to oscillations and simple harmonic motion
Introduction to oscillations and simple harmonic motionIntroduction to oscillations and simple harmonic motion
Introduction to oscillations and simple harmonic motionMichael Marty
 
Electrochemistry main (2)
Electrochemistry main (2)Electrochemistry main (2)
Electrochemistry main (2)Poonam Singh
 

Tendances (20)

Topic 1 shm
Topic 1 shmTopic 1 shm
Topic 1 shm
 
Physics
PhysicsPhysics
Physics
 
Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...
Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...
Synthesis of Bismuth Ferrite nano particles by sol-gel method and their chara...
 
Wave Optics Class 12 Part-1
Wave Optics Class 12 Part-1Wave Optics Class 12 Part-1
Wave Optics Class 12 Part-1
 
Chapter 7 nuclear physics
Chapter 7 nuclear physicsChapter 7 nuclear physics
Chapter 7 nuclear physics
 
Ionization energy
Ionization energyIonization energy
Ionization energy
 
Standing waves
Standing wavesStanding waves
Standing waves
 
nuclear binding energy
 nuclear binding energy nuclear binding energy
nuclear binding energy
 
Ch 31 Nuclear Physics and Radioactivity
Ch 31 Nuclear Physics and RadioactivityCh 31 Nuclear Physics and Radioactivity
Ch 31 Nuclear Physics and Radioactivity
 
General Properties of Nuclear
General Properties of NuclearGeneral Properties of Nuclear
General Properties of Nuclear
 
Chapter 6.1 : Introduction to Chemical Bonding
Chapter 6.1 : Introduction to Chemical BondingChapter 6.1 : Introduction to Chemical Bonding
Chapter 6.1 : Introduction to Chemical Bonding
 
Electrochemistry 12
Electrochemistry 12Electrochemistry 12
Electrochemistry 12
 
Chapter 21 Lecture- Nuclear Chemistry
Chapter 21 Lecture- Nuclear ChemistryChapter 21 Lecture- Nuclear Chemistry
Chapter 21 Lecture- Nuclear Chemistry
 
Wave functions
Wave functionsWave functions
Wave functions
 
Nuclear radiation
Nuclear radiationNuclear radiation
Nuclear radiation
 
IB Chemistry Limiting, Excess, Theoretical and Percentage Yield
IB Chemistry Limiting, Excess, Theoretical and Percentage YieldIB Chemistry Limiting, Excess, Theoretical and Percentage Yield
IB Chemistry Limiting, Excess, Theoretical and Percentage Yield
 
Class XII Solutions
Class XII   Solutions Class XII   Solutions
Class XII Solutions
 
Strongly Interacting Atoms in Optical Lattices
Strongly Interacting Atoms in Optical LatticesStrongly Interacting Atoms in Optical Lattices
Strongly Interacting Atoms in Optical Lattices
 
Introduction to oscillations and simple harmonic motion
Introduction to oscillations and simple harmonic motionIntroduction to oscillations and simple harmonic motion
Introduction to oscillations and simple harmonic motion
 
Electrochemistry main (2)
Electrochemistry main (2)Electrochemistry main (2)
Electrochemistry main (2)
 

En vedette (20)

Anschp21
Anschp21Anschp21
Anschp21
 
Anschp38
Anschp38Anschp38
Anschp38
 
Anschp17
Anschp17Anschp17
Anschp17
 
Anschp16
Anschp16Anschp16
Anschp16
 
Anschp33
Anschp33Anschp33
Anschp33
 
Anschp34
Anschp34Anschp34
Anschp34
 
Anschp26
Anschp26Anschp26
Anschp26
 
Anschp18
Anschp18Anschp18
Anschp18
 
Anschp19
Anschp19Anschp19
Anschp19
 
Anschp20
Anschp20Anschp20
Anschp20
 
Anschp36
Anschp36Anschp36
Anschp36
 
Anschp31
Anschp31Anschp31
Anschp31
 
Anschp25
Anschp25Anschp25
Anschp25
 
Anschp32
Anschp32Anschp32
Anschp32
 
Anschp35
Anschp35Anschp35
Anschp35
 
Anschp27
Anschp27Anschp27
Anschp27
 
Anschp30
Anschp30Anschp30
Anschp30
 
Anschp24
Anschp24Anschp24
Anschp24
 
Anschp29
Anschp29Anschp29
Anschp29
 
Anschp23
Anschp23Anschp23
Anschp23
 

Similaire à Anschp39

8m_ATOMS__NUCLEI.pdf chapter best notes preparation
8m_ATOMS__NUCLEI.pdf chapter best notes preparation8m_ATOMS__NUCLEI.pdf chapter best notes preparation
8m_ATOMS__NUCLEI.pdf chapter best notes preparation30jayporwal
 
revision xi - chapters1-5.pdf
revision xi - chapters1-5.pdfrevision xi - chapters1-5.pdf
revision xi - chapters1-5.pdfssuserfa137e1
 
Atomic structure
Atomic structureAtomic structure
Atomic structuresuresh gdvm
 
Atomic Structure ( sri chaitanya).pdf
Atomic Structure ( sri chaitanya).pdfAtomic Structure ( sri chaitanya).pdf
Atomic Structure ( sri chaitanya).pdfssuseree13e2
 
How to "see" a neutrino?
How to "see" a neutrino?How to "see" a neutrino?
How to "see" a neutrino?Alan Poon
 
Structure of atom
Structure of atom Structure of atom
Structure of atom sahil9100
 
Nuclei And Atoms Class 12
Nuclei And Atoms Class 12Nuclei And Atoms Class 12
Nuclei And Atoms Class 12Self-employed
 
Class 12th Physics Atom nuclei PPt
Class 12th Physics Atom nuclei PPtClass 12th Physics Atom nuclei PPt
Class 12th Physics Atom nuclei PPtArpit Meena
 
Notes for Atoms Molecules and Nuclei - Part III
Notes for Atoms Molecules and Nuclei - Part IIINotes for Atoms Molecules and Nuclei - Part III
Notes for Atoms Molecules and Nuclei - Part IIIEdnexa
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.pptssuser9306b4
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.pptdgzhfh
 
Atomic_Nucleus.ppt for general physics 2
Atomic_Nucleus.ppt for general physics 2Atomic_Nucleus.ppt for general physics 2
Atomic_Nucleus.ppt for general physics 2JosephMuez2
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.pptmragarwal
 

Similaire à Anschp39 (18)

8m_ATOMS__NUCLEI.pdf chapter best notes preparation
8m_ATOMS__NUCLEI.pdf chapter best notes preparation8m_ATOMS__NUCLEI.pdf chapter best notes preparation
8m_ATOMS__NUCLEI.pdf chapter best notes preparation
 
8m atoms _nuclei
8m atoms _nuclei8m atoms _nuclei
8m atoms _nuclei
 
Atoms and nuclei
Atoms and nucleiAtoms and nuclei
Atoms and nuclei
 
revision xi - chapters1-5.pdf
revision xi - chapters1-5.pdfrevision xi - chapters1-5.pdf
revision xi - chapters1-5.pdf
 
Atomic structure
Atomic structureAtomic structure
Atomic structure
 
Atomic Structure ( sri chaitanya).pdf
Atomic Structure ( sri chaitanya).pdfAtomic Structure ( sri chaitanya).pdf
Atomic Structure ( sri chaitanya).pdf
 
How to "see" a neutrino?
How to "see" a neutrino?How to "see" a neutrino?
How to "see" a neutrino?
 
Structure of atom
Structure of atom Structure of atom
Structure of atom
 
Nuclei And Atoms Class 12
Nuclei And Atoms Class 12Nuclei And Atoms Class 12
Nuclei And Atoms Class 12
 
Class 12th Physics Atom nuclei PPt
Class 12th Physics Atom nuclei PPtClass 12th Physics Atom nuclei PPt
Class 12th Physics Atom nuclei PPt
 
Notes for Atoms Molecules and Nuclei - Part III
Notes for Atoms Molecules and Nuclei - Part IIINotes for Atoms Molecules and Nuclei - Part III
Notes for Atoms Molecules and Nuclei - Part III
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.ppt
 
11 Nuclear
11 Nuclear11 Nuclear
11 Nuclear
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.ppt
 
Atomic_Nucleus.ppt for general physics 2
Atomic_Nucleus.ppt for general physics 2Atomic_Nucleus.ppt for general physics 2
Atomic_Nucleus.ppt for general physics 2
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.ppt
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.ppt
 
atomic_nucleus.ppt
atomic_nucleus.pptatomic_nucleus.ppt
atomic_nucleus.ppt
 

Anschp39

  • 1. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition Chapter 39. Nuclear Physics and the Nucleus The Elements 208 39-1. How many neutrons are in the nucleus of 82 Pb ? How many protons? What is the ratio N/Z? (N is the number of neutrons and Z is the number of protons.) A A = N + Z; N = A – Z = 126 neutrons; Z = 82 protons; = 1.54 Z 39-2. The nucleus of a certain isotope contains 143 neutrons and 92 protons. Write the symbol for this nucleus. 235 A = N + Z = 143 + 92 = 235; Z = 92: 92 U 39-3. From a stability curve it is determined that the ratio of neutrons to protons for a cesium nucleus is 1.49. What is the mass number for this isotope of cesium? N Z = 55; = 1.49; N = 1.49(55) = 81.95; A = N + Z ; A = 137 Z 39-4. Most nuclei are nearly spherical in shape and have a radius that may be approximated by 1 r = r0 A 3 r0 = 1.2 x 10−15 m 197 What is the approximate radius of the nucleus of a gold atom 79 Au ? r = (1.2 x10−15 m) 3 197 = 6.98 x 10-15 m ; r = 6.98 x 10-15 m 39-5. Study Table 39-4 for information on the several nuclides. Determine the ratio of N/Z for the following nuclides: Beryllium-9, Copper-64, and Radium 224. N Beryllium: A = 9; Z = 4; N = 9 – 4 = 5; = 1.25 Z 26
  • 2. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition N 39-5. (Cont.) Copper: A = 64; Z = 29; N = 64 – 29 = 35; = 1.21 Z N Radium: A = 224; Z = 88; N = 224 – 88 = 136; = 1.55 Z The Atomic Mass Unit 39-6. Find the mass in grams of a gold particle containing two million atomic mass units?  1.66 x 10-27 kg  m = 2 x 106 u  ; m = 3.32 x 10-21 kg  1.00 u  39-7. Find the mass of a 2-kg copper cylinder in atomic mass units? In Mev? In joules?  1u  m = 2 kg  -27 ; m = 1.20 x 1027 u  1.6606 x 10 kg   931 MeV  m = 1.204 x 1027 u  ; m = 1.12 x 1030 MeV  1u  30  1.6 x 10-13 J  m = 1.12 x 10 MeV  ; m = 1.79 x 1017 J  1 MeV  39-8. A certain nuclear reaction releases an energy of 5.5 MeV. How much mass (in atomic mass units) is required to produce this energy?  1u  m = 5.5 MeV  ; m = 0.00591 u  931 MeV  39-9. The periodic table gives the average mass of a silver atom as 107.842 u. What is the average mass of the silver nucleus? (Recall that me = 0.00055 u ) For silver, Z = 47. Thus, the nuclear mass is reduced by the mass of 47 electrons. m = 107.842 u – 47(0.00055 u); m = 107.816 u 27
  • 3. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-10. Consider the mass spectrometer as illustrated by Fig. 39-3. A uniform magnetic field of 0.6 T is placed across both upper and lower sections of the spectrometer., and the electric field in the velocity selector is 120 V/m. A singly charged neon atom (+1.6 x 10-19 C) of mass 19.992 u passes through the velocity selector and into the spectrometer. What is the velocity of the neon atom as it emerges from the velocity selector? E 120, 000 V/m v= = ; v = 2 x 105 m/s B 0.6 T R *39-11. What is the radius of the circular path followed by the neon atom of Problem 19-10? B = 0.6 T into paper  1.66 x 10-27 kg  −26 m = 19.992 u   = 3.32 x10 kg  1u  mv 2 mv (3.32 x 10-27 kg)(2 x 105 m/s) = qvB; R= ; R= ; R = 6.92 cm R qB (1.6 x 10-19 C)(0.600 T) Mass Defects and Binding Energy (Refer to Table 39-4 for Nuclidic Masses.) 20 *39-12. Calculate the mass defect and binding energy for the neon-20 atom 10 Ne . mD = [( ZmH + Nmn )] − M = [ 10(1.007825 u) + 10(1.008665 u)] − 19.99244 u mD = 20.016490 u –19.99244 u ; mD = 0.17246 u  931 MeV  E = mD c 2 = (0.17246 u)   = 160.6 MeV ; E = 161 MeV  1u  3 *39-13. Calculate the binding energy and the binding energy per nucleon for tritium 1 H . How much energy in joules is required to tear the nucleus apart into its constituent nucleons? mD = [( ZmH + Nmn )] − M = [ 1(1.007825 u) + 2(1.008665 u) ] − 3.016049 u 28
  • 4. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition  931 MeV  39-13. (Cont.) mD = 0.009106 u; E = mD c 2 = (0.009106 u)   = 8.48 MeV  1u  EB 8.48 MeV EB E = 8.48 MeV; = ; = 2.83 MeV/nucleon A 3 A Energy to tear apart: EB = 8.48 MeV(1.6 x 10-13 J/MeV) = 1.36 x 10-12 J *39-14. Calculate the mass defect of 7 Li . What is the binding energy per nucleon? 3 mD = [( ZmH + Nmn )] − M = [ 3(1.007825 u) + 4(1.008665 u) ] − 7.016930 u mD = 7.058135 u –7.016930 u ; mD = 0.041205 u  931 MeV  E = mD c 2 = (0.041205 u)   = 38.4 MeV ; E = 38.4 MeV  1u  EB 38.4 MeV EB = ; = 5.48 MeV/nucleon A 7 A 12 *39-15. Determine the binding energy per nucleon for carbon-12 ( 6 C ). mD = [ 6(1.007825 u) + 6(1.008665 u) ] − 12.0000 u = 0.09894 u  931 MeV  EB 92.1 MeV E = mD c 2 = (0.09894 u)  ; = = 7.68 MeV/nucleon  1u  A 12 197 *39-16. What is the mass defect and the binding energy for a gold atom 79 Au ? mD = [ 79(1.007825 u) + 118(1.008665 u) ] − 196.966541 u = 1.674104 u  931 MeV  E = mD c 2 = (1.674104 u)   = 1.56 GeV;  1u  EB 1.56 GeV = = 7.91 MeV/nucleon A 197 29
  • 5. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 120 *39-17. Determine the binding energy per nucleon for tin-120 ( 50 Sn ). mD = [ 50(1.007825 u) + 70(1.008665 u) ] − 119.902108 u = 1.09569 u EB  (1.09569 u)(931 MeV/u  EB = ; = 8.50 MeV/nucleon A  120  A Radioactivity and Nuclear Decay 39-18. The activity of a certain sample is rated as 2.8 Ci. How many nuclei will have disintegrated in a time of one minute?  3.7 x 1010s-1  Nuclei = 2.8 Ci   (60 s); nuclei = 6.22 x 1012 nuclei  1 Ci  60 39-19. The cobalt nucleus 27 Co emits gamma rays of approximately 1.2 MeV. How much mass is lost by the nucleus when it emits a gamma ray of this energy?  1u  E = 1.2 MeV  ; m = 0.00129 u  931 MeV  39-20. The half-life of the radioactive isotope indium-109 is 4.30 h. If the activity of a sample is 1 mCi at the start, how much activity remains after 4.30, 8.60, and 12.9 h? t / T½ 1 1 t 4.3 h 1 R = R0   ; = = 1; R = (1 mCi)   ; R = 0.5 mCi 2 T½ 4.3 h 2 t / T½ 2 1 t 8.6 h 1 R = R0   ; = = 2; R = (1 mCi)   ; R = 0.25 mCi 2 T½ 4.3 h 2 t / T½ 3 1 t 12.9 h 1 R = R0   ; = = 3; R = (1 mCi)   ; R = 0.125 mCi 2 T½ 4.3 h 2 30
  • 6. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 39-21. The initial activity of a sample containing 7.7 x 1011 bismuth-212 nuclei is 4.0 mCi. The half-life of this isotope is 60 min. How many bismuth-212 nuclei remain after 30 min? What is the activity at the end of that time? t / T½ ½ 1 t 30 min 1 N = N0   ; = = 0.5; N = (7.7 x 1011 )   = 5.44 x 1011 nucl. 2 T½ 60 min 2 t / T½ ½ 1 1 R = R0   ; R = (4 mCi)   ; R = 2.83 mCi 2 2 *39-22. Strontium-90 is produced in appreciable quantities in the atmosphere during a nuclear explosion. If this isotope has a half-life of 28 years, how long will it take for the initial activity to drop to one-fourth of its original activity? n n 1 R 1 1 R = R0   ; =  = ; n=2 2 R0  2  4 t t n= = = 2; t = 2(28 yr); t = 56 yr T½ 28 yr *39-23. Consider a pure, 4.0-g sample of radioactive Gallium-67. If the half-life is 78 h, how much time is required for 2.8 g of this sample to decay? When 2.8 g decay, that leaves 4 g – 2.8 g or 1.20 g remaining. n n 1 m 1.2 g 1 m = m0   ; = = 0.300;   = 0.300 2 m0 4g 2 The solution is accomplished by taking the common log of both sides: −0.523 n log(0.5) = log(0.300); − 0.301n = −0.523; n= = 1.74 −0.301 t t n= = = 1.74; t = 1.74 (78 h) t = 135 h T½ 78 h 31
  • 7. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-24. If one-fifth of a pure radioactive sample remains after 10 h, what is the half-life? n 1 1 =  ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699 5 2 t 10 h 10 h n= = = 2.32; T½ = ; T½ = 4.31 h T½ T½ 2.32 Nuclear Reactions *39-25. Determine the minimum energy released in the nuclear reaction 19 9 F +1 H →4 He + 16 O + energy 1 2 8 The atomic mass of 19 9 F is 18.998403 u, 4 2 He = 4.002603 u, 1 1H = 1.007824 u. E= 19 9 F + 1H - 2 He - 16 O ; (Energy comes from mass defect) 1 4 8 E = 18.998403 u + 1.007825 u – 4.002603 – 15.994915 u = 0.0087 u  931 MeV  E = (0.00871 u)  ; E = 8.11 MeV  1u  *39-26. Determine the approximate kinetic energy imparted to the alpha particle when Radium-226 decays to form Radon-222. Neglect the energy imparted to the radon nucleus. 226 88 Ra → 222 86 Rn + 4 He + Energy; 2 226 88 Ra = 226.02536  931 MeV  E = 226.02536 u - 222.017531 u - 4.002603 u = 0.00523 u   = 4.87 MeV  1u  *39-27. Find the energy involved in the production of two alpha particles in the reaction 7 3 Li + 1H → 2 He + 2 He + energy 1 4 4 32
  • 8. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition  931 MeV  E = 7.016003 u + 1.007825 u - 2(4.002603 u) = 0.018622 u   = 17.3 MeV  1u  *39-28. Compute the kinetic energy released in the beta minus decay of thorium-233. 233 90 Th → 233 91 Pa + 0 β + energy; +1 233 90 Th = 233.041469 u; 233 91 Pa = 233.040130 E = 233.041469 u - 233.040130 u - 0.00055 u = 0.000789 u  931 MeV  E = 0.000789 u  ; E = 0.735 MeV  1u  *39-29. What must be the energy of an alpha particle as it bombards a Nitrogen-14 nucleus 17 producing 8 O and 1H? (17 O = 16.999130 u) . 1 8 4 2 He + 14 N + Energy → 7 17 8 1 O + 1H  931 MeV  E = 16.999130 u + 1.007825 u - 14.003074 - 4.002603 u = 0.001278 u    1u  E = 1.19 MeV This is the Threshold Energy for the reaction Challenge Problems *39-30. What is the average mass in kilograms of the nucleus of a boron-11 atom?  1.66 x 10-27 kg  m = 11.009305 u  ; m = 1.83 x 10-26 kg  1.00 u  *39-31. What are the mass defect and the binding energy per nucleon for boron-11? mD = [( ZmH + Nmn )] − M = [ 5(1.007825 u) + 6(1.008665 u) ] − 11.009305 u mD = 11.09112 u –11.009305 u ; mD = 0.08181 u  931 MeV  E = mD c 2 = (0.08181 u)   = 76.2 MeV ; E = 76.2 MeV  1u  33
  • 9. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition EB 76.2 MeV EB = ; = 6.92 MeV/nucleon A 11 A *39-32. Find the binding energy per nucleon for Thallium-206. mD = [ 81(1.007825 u) + 125(1.008665 u) ] − 205.976104 u = 1.740846 u  931 MeV  EB 1621 MeV E = mD c 2 = (1.740846 u)  ; = = 7.87 MeV/nucleon  1u  A 206 *39-33. Calculate the energy required to separate the nucleons in mercury-204. mD = [ 80(1.007825 u) + 124(1.008665 u) ] − 203.973865 u = 1.7266 u  931 MeV  E = mD c 2 = (1.7266 u)  ; EB = 1610 MeV  1u  *39-34. The half-life of a radioactive sample is 6.8 h. How much time passes before the activity drops to one-fifth of its initial value? n R 1 1 = =  ; 0.200 = (0.5) n ; n log(0.5) = log(0.2); − 0.301 n = −0.699 R0 5  2  0.699 t 6.8 h 6.8 h n= = 2.32; n = = = 2.32; T½ = ; T½ = 2.93 h 0.301 T½ T½ 2.32 *39-35. How much energy is required to tear apart a deuterium atom? (1 H = 2.014102 u) 2 mD = [ 1(1.007825 u) + 1(1.008665 u) ] − 2.014102 u = 0.002388 u  931 MeV  E = mD c 2 = (0.002388 u)  ; EB = 2.22 MeV  1u  34
  • 10. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition *39-36. Plutonium-232 decays by alpha emission with a half-life of 30 min. How much of this substance remains after 4 h if the original sample had a mass of 4.0 g? Write the equation for the decay? t / T½ 8 1 t 4h 1 m = m0   ; = = 8.00; m = (4 g)   ; m = 15.6 mg 2 T½ 0.5 h 2 232 94 Pu → 228 92 U + 4 He + energy 2 *39-37. If 32 x 109 atoms of a radioactive isotope are reduced to only 2 x 109 atoms in a time of 48 h, what is the half-life of this material? n N 2 x 109  1  = 9 =   ; 0.0625 = (0.5) n ; n log(0.5) = log(0.0625); − 0.301 n = −1.204 N 0 32 x 10  2  1.204 t 48 h 48 h n= = 4.00; n = = = 4.00; T½ = ; T½ = 12.0 h 0.301 T½ T½ 4.00 *39-38. A certain radioactive isotope retains only 10 percent of its original activity after a time of 4 h. What is the half-life? n N 1 1 = =   ; 0.100 = (0.5) n ; n log(0.5) = log(0.100); − 0.301 n = −1.00 N 0 10  2  1.00 t 4h 4h n= = 3.32; n = = = 3.32; T½ = ; T½ = 72.3 min 0.301 T½ T½ 3.32 *39-39. When a 6 Li nucleus is struck by a proton, an alpha particle and a produce nucleus are 3 released. Write the equation for this reaction. What is the net energy transfer in this case? 6 3 Li + 1H → 2 He + 3 He + energy 1 4 2 35
  • 11. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition  931 MeV  E = 6.015126 u + 1.007825 u - 4.002603 u - 3.016030 = 0.00432 u   = 4.02 MeV  1u  *39-40. Uranium-238 undergoes alpha decay. Write the equation for the reaction and calculate the disintegration energy. 238 92 U→ 224 90 4 Th + 2 He + energy E = 238.05079 u - 234.04363 u - 4.002603 u = 0.00456 u  931 MeV  E = 0.00456 u  ; E = 4.24 MeV  1u  *39-41. A 9-g sample of radioactive material has an initial activity of 5.0 Ci. Forty minutes later, the activity is only 3.0 Ci. What is the half-life? How much of the pure sample remains? n R 3 Ci  1  = =  ; 0.600 = (0.5) n ; n log(0.5) = log(0.6); − 0.301 n = −0.222 R0 5 Ci  2  0.222 t 40 min 40 min n= = 0.737; n = = = 0.737; T½ = ; T½ = 54.3 min 0.301 T½ T½ 0.737 n 0.737 1 1 m = m0   = (9 g)   m = 5.40 g 2 2 Critical Thinking Problems *39-42. Nuclear fusion is a process that can produce enormous energy without the harmful byproducts of nuclear fission. Calculate the energy released in the following nuclear fusion reaction: 3 2 H + 3 He →2 He +1 H+1H + energy 2 4 1 1  931 MeV  E = 2(3.016030 u) - 4.002603 u - 2(1.007825 u) = 0.013807 u    1u  36
  • 12. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition E = 12.9 MeV *39-43. Carbon-14 decays very slowly with a half-life of 5740 years. Carbon dating can be accomplished by seeing what fraction of carbon-14 remains, assuming the decay process began with the death of a living organism. What would be the age of a chunk of charcoal if it was determined that the radioactive C-14 remaining was only 40 percent of what would be expected in a living organism? n R 1 = 0.40 =   ; 0.400 = (0.5) n ; n log(0.5) = log(0.4); − 0.301 n = −0.399 R0 2 0.399 t t n= = 0.737; n = = = 1.32; t = 1.32(5740 yr); t = 7590 yr 0.301 T½ 5740 yr *39-44. The velocity selector in a mass spectrometer has a magnetic field of 0.2 T perpendicular to an electric field of 50 kV/m. The same magnetic field is across the lower region. What is the velocity of singly charged Lithium-7 atoms as they leave the selector? If the radius of the path in the spectrometer is 9.10 cm, find the atomic mass of the lithium atom? E 50, 000 V/m v= = ; v = 2.5 x 105 m/s B 0.2 T mv 2 qBR (1.6 x 10-19 C)(0.2 T)(0.091 m) R = qvB; m = ; m= ; R v (2.5 x 105 m/s)  1u  m = 1.165 x 10-26 kg  -27 ; m = 7.014 u B = 0.2 T into paper  1.66 x 10 kg  *39-45. A nuclear reactor operates at a power level of 2.0 MW. Assuming that approximately 200 MeV of energy is released for a single fission of U-235, how many fission processes are occurring each second in the reactor? 37
  • 13. Chapter 39. Nuclear Physics and the Nucleus Physics, 6th Edition 2 x 106 J/s 1.25 x 1019 MeV/s P= = 1.25 x 1019 MeV/s; = 6.25 x 10216 fissions/s 1.6 x 10-19 J/MeV 200 MeV/fission 14 *39-46. Consider an experiment which bombards 7 N with an alpha particle. One of the two product nuclides is 1 H . The reaction is 1 4 2 He +14 N → Z X +1 H 7 A 1 What is the product nuclide indicated by the symbol X? How much kinetic energy must the alpha particle have in order to produce the reaction? Conservation of nucleons means: 4 2 He +14 N + enery → 7 17 8 O +1 H 1 E = 16.99913 u + 1.007825 u - 4.002603 u - 14.003074 u = 0.001282 u  931 MeV  E = 0.001282 u  ; E = 1.19 MeV  1u  *39-47. When passing a stream of ionized lithium atoms through a mass spectrometer, the radius of the path followed by 7 Li (7.0169 u) is 14.00 cm. A lighter line is formed by the 6 Li 3 3 6 (6.0151 u). What is the radius of the path followed by the 3 Li isotopes? mv 2 mv = qvB; R = ; (See Problems 39-10 and 39-11) R qB m1 R1 m2 R1 (6.0151 u)(14 cm) = ; R2 = = ; R2 = 6.92 cm m2 R2 m1 7.0169 u 38