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Understanding Nuclear Energy Transformations and Reactions

Discover the source and transformations of nuclear energy, magic numbers in nuclei, and calculations behind nuclear reactions and fission. Explore the principles and hypotheses of nuclear forces and density, with practical examples and calculations.

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Understanding Nuclear Energy Transformations and Reactions

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  1. What is the source of Nuclear Energy? Which sorts of nuclear transformations will yield energy?

  2. Magic numbers of n, p – separately:2, 8, 20, 28, 50, 82, 126, …

  3. Mass (6,12)C =12.0… u or 12x931.5 MeV • ------------------------------------------------------------- • Mass (1,1) H (p+e) = 1.007825 u • Mass (0,1) n = 1.008665 u • Mass (1,2) D = 2.014102 u • Diff = 0.002388u = 2.224 Mev or 1.112 Mev/N • --------------------------------------------------------------- • 26 H +30 n is 56.46340 u • (26,56)Fe is 55.93494 u • Diff 0.52846 u = 492.26 MeV or • 8.79 Mev per N

  4. How the plot is made is previous slide. What does that shape mean? • Hypothesis: Nuclear force is strong but short range. It’s reach is the diameter of a nucleus a little to the left of the peak. • Useful fact: Nuclear density is constant – like apples in a box. The volume filled is the volume of the apples – they don’t compress when packed. The radius of a nucleus is r = 1.2 A^(1/3) f, (1 f =10^-15 m) • v=(4/3)pi r^3 or v=((4/3) 1.2)^3 A

  5. A r (f) element • 8 2.4 Lithium or Berylium • 27 3.6 Aluminum • 64 4.8 Zinc • 125 6.0 Antimony or Telurium • 216 7.2 Astatine (Lead is 208)

  6. n + (92,235)U  (56,141)Ba +(36,92)Kr + 3n • U 235.043915 u Ba 140.9139 u • n 1.008665 u Kr 91.8973 u • 3n 3.0260 u • Sum 236.052580 u 235.8372 u • Diff = 0.2154 u = 200.6 MeV • Fr dif = 0.000916 • 1 Kg (92,235)U loses 0.000916 Kg if all fission • Mc^2 = 0.000916 x (3e8)^2 = 8.247e13 J = 23 Mkwh. 1600 kwh/mo (my House) • 14,000 homes for 1 month

  7. How to make it happen? Use a neutron.

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