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Nuclear Chemistry and Mass-Energy Relationships

Nuclear Chemistry and Mass-Energy Relationships. Chapter 3 – Part 2. Mass Excess or Mass Defect. Δ = M – A = nuclidic mass – mass number Example Calculate mass excess in MeV for 14 C. The nuclidic mass of 14 C is 14.00324 daltons.

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Nuclear Chemistry and Mass-Energy Relationships

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  1. Nuclear Chemistry and Mass-Energy Relationships Chapter 3 – Part 2

  2. Mass Excess or Mass Defect • Δ= M – A = nuclidic mass – mass number Example • Calculate mass excess in MeV for 14C. The nuclidic mass of 14C is 14.00324 daltons. Δ= M – A = 14.00324 – 14 = 3.24 x 10-3 daltons = 3.24 x 10-3 daltons x 931.5 MeV/dalton = 3.02 MeV

  3. Energy Changes in Nuclear Reactions • Q-value • Q = (Σmassesreactants-Σmassesproducts) x 931.5 MeV/daltons • Q = Σ Δreactants - Σ Δproducts • Q-value calculator http://www.nndc.bnl.gov/qcalc/ • Atomic Mass Data Centerhttp://www.nndc.bnl.gov/amdc/

  4. Q-value Energy released in a nuclear reaction (> 0 if energy is released, < 0 if energy is used) Example: The sun is powered by the fusion of hydrogen into helium: 4p  4He + 2 e+ + 2ne Mass difference dMreleased as energydE = dM x c2

  5. Binding Energy • E (MeV) = Masses of reactants – Masses of products) 931.5 MeV/dalton BE = {Zmp + Nmn – [m(AX) –Zme]}c2 if using atomic masses - BE = {ZmH + Nmn – m(AX)}c2 (Remember 1 u = 1 dalton = 931.5 MeV/c2

  6. Binding Energy Energy that is released when a nucleus is assembled from neutrons and protons mp = proton mass, mn = neutron mass, m(Z,N) = mass of nucleus with Z,N • B = 0 for H, otherwise B > 0 • 2D1 - deteriumBE = (1.007825 + 1.008665 - 2.0141) x 931.481 MeV = 2.226 MeV • 4He2BE = (2*1.007825 + 2*1.008665 - 4.002603) x 931.481 MeV = 28.30 MeV • 238U146BE = (92*1.007825 + 146*1.008665 - 238.0289) x 931.481 MeV = 1822.06 MeV The more nucleons packed into a nucleus, the more energy is released, and thus the higher the binding energy.

  7. Binding Energy per Nucleon Source: http://hyperphysics.phy-astr.gsu.edu/

  8. Energy Changes in Radioactive Decay • Alpha Decay 252Cf  248Cm + α Q = ΔCf – (ΔCm +Δα) = 76.030 – (2.424 + 67.388) = 6.22 MeV

  9. Energy Changes in Radioactive Decay • Negatron Decay 32P  32S + e- + antineutrino + Q Q = ΔP- ΔS = -24.305 – (-26.016) = 1.711 MeV • Positron Decay 26Al  26Mg + e+ + atomic electron + neutrino + Q Q = Δ(26Al) – [Δ(26Mg) + 2me] = = -12.210 – (- 16.214 – 2(0.511)) = 2.982 MeV

  10. QPuU + QUNp = QPu  Am + QAm Np QUNp = QPu  Am + QAm Np- QPuU QUNp = 0.0208 + 5.49 -4.90 = 0.61 MeV α, 5.49 MeV 241Am 237Np β-, 0.0208 MeV β- α, 4.90 MeV 241Pu 237U Closed-Cycle Decay for Mass-Energy Calculations

  11. Semiempirical Binding Energy Equation a) Volume BE/A ~ 8 MeV, so BE ≈ 8A or BEA  avA b) Surface nuclei on surface have fewer neighbors – volume term will over-estimate BE surface area of sphere 4πR2 so R2  (r0A1/3)2 A2/3 – asA2/3 c) Coulomb Coulomb repulsion of protons – each proton repels all others Z protons repelling (Z – 1) protons estimate using Coulomb energy for a sphere – Collect constants into one ac = 0.72 MeV – acZ(Z-1)A-1/3

  12. Semiempirical Binding Energy Equation d) Symmetry Light stable nuclei have N≈Z, heavy nuclei have N>Z Too many protons, unstable; too many neutrons, unstable – asym(A - 2Z)2/A e) Pairing Preferred BE for even Z, N Pairing energy δ δ = +apA-3/4 even Z and even N δ = 0 odd Z, even N OR even Z, odd N δ = -apA-3/4 odd Z and odd N

  13. Binding Energy Best fit values (from A.H. Wapstra, Handbuch der Physik 38 (1958) 1) in MeV/c2

  14. Nuclear Energy Surface Diagram For a constant A Binding energy per nucleon along const A due to asymmetry term in mass formula 2δ-displacement valley of stability (Bertulani & Schechter)

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