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Nuclear and Particle Physics

Nuclear and Particle Physics. Nuclear Physics. Back to Rutherford and his discovery of the nucleus Also coined the term “proton” in 1920, and described a “neutron” in 1921 Neutron discovered by Chadwick in 1932. Ernest Rutherford 1871-1937. m e = 9.1 x 10 -31 kg

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Nuclear and Particle Physics

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  1. Nuclear and Particle Physics

  2. Nuclear Physics Back to Rutherford and his discovery of the nucleus Also coined the term “proton” in 1920, and described a “neutron” in 1921 Neutron discovered by Chadwick in 1932 Ernest Rutherford 1871-1937 me = 9.1 x 10-31 kg mN = 1.6749 x 10-27 kg mP = 1.6726 x 10-27kg nucleons James Chadwick 1891-1974

  3. Nuclides and Isotopes To specify a nuclide: Z is the atomic number = number of electrons or protons A is the mass number = number of neutrons + protons So number of neutrons = A-Z Number of protons = Z Isotopes – same atomic number, different mass number e.g. carbon: Many isotopes do not occur naturally, also elements > U

  4. Sizes We saw with the Bohr model that radius of the atom depended on atomic number Nucleus = protons + neutrons = mass number The volume of a nucleus is proportional to the mass number

  5. Masses Mass spectrometer 1 atomic mass unit (u.) = 1.6606 x 10-27 kg = 931.5 MeV Fixed so that carbon = 12.00000 u mN = 1.6749 x 10-27 kg = 1.0087 u mP = 1.6726 x 10-27kg = 1.0078 u

  6. Binding Energy Total mass of a nucleus < sum of masses Example: Mass of helium nucleus = 6.6447 x 10-27 kg Contains 2 protons and 2 neutrons Mass = 2 x (1.6749 x 10-27 + 1.6726 x 10-27 ) kg = 6.6950 x 10-27 kg Difference = (6.6950 – 6.6447) x 10-27 = 0.0503 x 10-27 kg Energy = mc2 = 0.0503 x 10-27 x c2 = 4.53 x 10-12 J = (4.53 x 10-12) / (1.6 x 10-19) = 2.83 x 107 eV = 28.3 MeV

  7. Atomic Mass Units 1 u = 931.5 MeV mN = 1.6749 x 10-27 kg = 1.0087 u mP = 1.6726 x 10-27kg = 1.0078 u Mass of helium nucleus = 4.0026 u

  8. Atomic Mass Units Same calculation Mass of 2p + 2n = 2 x (1.0078 + 1.0087) = 4.0330 u 4.0330 u Difference = 0.0304 u Binding energy = 0.0305 x 931.5 = 28.3 MeV

  9. Average Binding Energy He – 4 nucleons, 28.3 MeV total: average = 7.075MeV • Graph

  10. Attractive? How does nucleus stay together? Like charges repel! Force stronger than electric force Strong nuclear force Short range (~10-15 m) Stable nuclides N = Z A > 30-40 – more neutrons Z > 82 – no stable nuclides Strong force can’t overcome repulsion

  11. Radioactivity Becquerel, 1896 Emission of radiation without external stimulus Curies – polonium (Po) and radium (Ra) Henri Becquerel 1852-1908 Radioactivity unaffected by heating, cooling, etc. 1903 (Physics) 1911 (Chem) Marie Curie 1867 - 1934 Pierre Curie 1859 - 1906

  12. Classification Rutherford classified 3 types of radioactivity according to penetration power Also different charge Video: “People Pretending to be Alpha Particles” Important factor: Conservation of nucleon number (neutrons + protons) = (neutrons + protons)

  13. Alpha Decay Least penetrating – nucleus of Radium 226 is an alpha emitter: transmutation Parent Daughter Mass of parent > mass of daughter + mass of alpha Difference = kinetic energy

  14. Example 232.03714 u  228.02873 u + 4.002603 u total = 232.03133 u Lost mass = 232.03714 – 232.03133 = 0.00581 u 0.00581u x 931.5 MeV/u = 5.4 MeV (some recoil)

  15. Beta decay  One electron What is lost is NOT an orbital electron Instead a neutron changes to a proton + electron So (6p + 8n) => (7p + 7n) + e-- decay

  16. Example Keep track of electrons! Carbon 14 has m = 14.003242 u 6 electrons Nitrogen 14 has m = 14.003074 u normally 7 electrons But in the decay, the nitrogen would have 6 electrons However the total on the r.h.s. of the equation has 7 So difference = 0.000168 u = 0.156 MeV = 156 keV

  17. Conservation of energy • Energy of decay = 156 keV = problem! ?

  18. A new particle • Proposed by Pauli (1930) - neutrino • Theory by Fermi • Discovered 1956 • Zero charge, ~0 rest mass Wolfgang Pauli 1900-1958 antineutrino Cosmic neutrino detection “Zero rest mass” – speed of light 1998 – Super Kamiokande – some mass Enrico Fermi 1901-1954

  19. More on positrons Many isotopes have more neutrons than protons • Decay by emission of electron Other isotopes have more protons than neutrons • Decay by emission of positron Proton changes to a neutron + positron + decay

  20. Annihilation Proton changes to a neutron + positron + decay Positron annihilation Application – positron emission tomography

  21. Positron Emission Tomography PET – basis – use radio-labelled compounds, i.e. those containing a radionuclide. Positron emitters: As an example, oxygen-15 can be used to look at oxygen metabolism and blood flow. Fluorine-18 is commonly used to examine cancerous tumours.

  22. PET - method Annihilation produces two back-to-back 511 keV photons Simultaneous detection

  23. Electron capture • Nucleus absorbs orbiting electron Proton changes to neutron Usually K electron X-ray emission as outer electron jumps down to K

  24. Gamma decay Most penetrating  = photon. High energy *Excited nucleus  lower energy state - (9.0 MeV) - (13.4 MeV) (4.4 MeV) Energy levels far apart = keV or MeV

  25. Homework . . . • p.902,#6; • p.908, Practice 25B; • p.912,Section Review • p.928, 30-37; • p. 930, 56,60; • Read through lab for next time; answer pre-lab questions

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