1251_Physics ProblemsTechnical Physics

1251_Physics ProblemsTechnical Physics - 592 P44.71 Nuclear...

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592 Nuclear Structure P44.71 For an electric charge density ρ π = Ze R 43 3 bg . Using Gauss’s Law inside the sphere, Er r Ze R ⋅= 4 2 3 0 3 : E Zer R = 1 4 0 3 rR af E Ze r = 1 4 0 2 We now find the electrostatic energy: UE r d r r =∈ = z 1 2 4 0 22 0 U Zer R rd r Ze r r R R R R R R F H G I K J +∈ F H G I K J = + L N M O Q P = zz 1 2 1 4 4 1 2 1 4 4 8 5 1 3 20 0 0 2 222 6 2 0 0 0 2 4 2 0 5 6 0 P44.72 (a) For the electron capture, 43 93 1 0 Tc e Mo 42 93 +→ + γ . The disintegration energy is QM M c =− 93 93 2 Tc Mo . Q = > 92 910 2 92 906 8 3 17 2 44 .. . . u 931.5 MeV u MeV MeV Electron capture is allowed to all specified excited states in 42 93 Mo. For positron emission, 43 93 1 0 Tc Mo e 42 93 →+ + + . The disintegration energy is ′= M
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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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