2020-05-04T17:06:48-04:00
Calculate the mass defect and nuclear binding energy per nucleon of each of the nuclides
a.O-16(atomic mass=15.994915 amu)
b.Ni-58(atomic mass=57.935346 amu)
c.Xe-129(atomic mass=15.994915 amu)
1
2020-05-05T18:40:55-0400
a)
Z = 8 , N = A − Z = 16 − 8 = 8 Z=8, N=A-Z=16-8=8 Z = 8 , N = A − Z = 16 − 8 = 8
Δ m = 8 ( 1.00783 ) + 8 ( 1.00866 ) − 15.994915 = 0.137005 a m u \Delta m=8(1.00783)+8(1.00866)-15.994915\\=0.137005\ amu Δ m = 8 ( 1.00783 ) + 8 ( 1.00866 ) − 15.994915 = 0.137005 am u
E = ( 0.137005 ) 931.5 16 = 7.976 M e V n u c l e o n E=(0.137005)\frac{931.5}{16}=7.976\frac{MeV}{nucleon} E = ( 0.137005 ) 16 931.5 = 7.976 n u c l eo n M e V b)
Z = 28 , N = A − Z = 58 − 28 = 30 Z=28, N=A-Z=58-28=30 Z = 28 , N = A − Z = 58 − 28 = 30
Δ m = 28 ( 1.00783 ) + 30 ( 1.00866 ) − 57.935346 = 0.534694 a m u \Delta m=28(1.00783)+30(1.00866)-57.935346 \\=0.534694\ amu Δ m = 28 ( 1.00783 ) + 30 ( 1.00866 ) − 57.935346 = 0.534694 am u
E = ( 0.534694 ) 931.5 58 = 8.732 M e V n u c l e o n E=(0.534694)\frac{931.5}{58}=8.732\frac{MeV}{nucleon} E = ( 0.534694 ) 58 931.5 = 8.732 n u c l eo n M e V
c)
Z = 54 , N = A − Z = 129 − 54 = 75 Z=54, N=A-Z=129-54=75 Z = 54 , N = A − Z = 129 − 54 = 75
Δ m = 54 ( 1.00783 ) + 75 ( 1.00866 ) − 128.904780 = 1.16754 a m u \Delta m=54(1.00783)+75(1.00866)-128.904780\\=1.16754\ amu Δ m = 54 ( 1.00783 ) + 75 ( 1.00866 ) − 128.904780 = 1.16754 am u
E = ( 1.16754 ) 931.5 129 = 8.431 M e V n u c l e o n E=(1.16754)\frac{931.5}{129}=8.431\frac{MeV}{nucleon} E = ( 1.16754 ) 129 931.5 = 8.431 n u c l eo n M e V
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