1234_Physics ProblemsTechnical Physics

1234_Physics ProblemsTechnical Physics - Chapter 44 Section...

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Chapter 44 575 Section 44.2 Nuclear Binding Energy P44.14 Using atomic masses as given in Table A.3, (a) For 1 2 H: −+ + 2 014 102 1 1 008 665 1 1 007 825 2 .. . bg E A b = F H G I K J = 0001194 931 5 111 . . . u MeV u MeV nucleon . (b) For 2 4 He: 2 1 008 665 2 1 007 825 4 002 603 4 . +− E A c b == 000759 707 2 u M e V n u c l e o n . (c) For 26 56 Fe: 30 1 008 665 26 1 007 825 55 934 942 0 528 . . = u E A c b = 0528 56 000944 879 2 . M e V n u c l e o n . (d) For 92 238 U: 146 1 008 665 92 1 007 825 238 050 783 1 934 2 . . = u E A c b = 19342 238 000813 757 2 . M e V n u c l e o n . P44.15 MZ m N m M =+− Hn BE A M A = 931 5 . af Nuclei ZNM in u M in u BE A in MeV 55 Mn 25 30 54.938 050 0.517 5 8.765 56 Fe 26 30 55.934 942 0.528 46 8.790 59 Co 27 32 58.933 200 0.555 35 8.768 56 Fe has a greater BE A than its neighbors. This tells us finer detail than is shown in Figure 44.5. P44.16 Use Equation 44.2. The 11 23 Na , E A b = 811 . MeV nucleon and for 12 23 Mg , E A b = 790 . M e Vn u c l e
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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