{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1234_Physics ProblemsTechnical Physics

# 1234_Physics ProblemsTechnical Physics - Chapter 44 Section...

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online