1 / 14

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.

Download Presentation

Nuclear Chemistry and Mass-Energy Relationships

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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)

More Related