1235_Physics ProblemsTechnical Physics

1235_Physics ProblemsTechnical Physics - 576 P44.18 Nuclear...

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576 Nuclear Structure P44.18 (a) The radius of the 40 Ca nucleus is: RrA == × = × −− 0 13 15 15 1 20 10 40 4 10 10 .. m m ej a f . The energy required to overcome electrostatic repulsion is U kQ R e ×⋅ × × = 3 5 3 8 99 10 20 1 60 10 5 4 10 10 135 10 841 2 91 9 2 15 11 . Nm C C m J M e V 22 e j . (b) The binding energy of 20 40 Ca is E b =+ = 20 1 007 825 20 1 008 665 39 962 591 931 5 342 . . u u u MeV u MeV bg b g . (c) The nuclear force is so strong that the binding energy greatly exceeds the minimum energy needed to overcome electrostatic repulsion. P44.19 The binding energy of a nucleus is EZ M N m M bn Z A MeV H X MeV u af a f 931 494 For 8 15 O: E b = 8 1 007 825 7 1 008 665 15 003 065 931 494 111 96 . . . u u u MeV u MeV b g . For 7 15 N: E b = 7 1 007 825 8 1 008 665 15 000 109 931 494 115 49 . . . u u u MeV u MeV b g . Therefore, the binding energy of N is larger by 3.54 MeV 7 15 . P44.20 Removal of a neutron from
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