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Nuclear Physics Chapter 25

Nuclear Physics Chapter 25. 3 Li lithium 6.941 1. atomic number (# of p + ). symbol. name. average atomic mass. electrons in outer energy level. in the nucleus…. Protons. positive charge. made of two up quarks and a down quark. Electric charge +2/3 e. Electric charge -1/3 e.

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Nuclear Physics Chapter 25

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  1. Nuclear Physics Chapter 25

  2. 3 Li lithium 6.941 1 atomic number (# of p+) symbol name average atomic mass electrons in outer energy level

  3. in the nucleus… Protons positive charge made of two up quarks and a down quark Electric charge +2/3 e Electric charge -1/3 e Neutrons made of two down quarks and an up quark

  4. = proton = neutron Li 7 Li 6 number of nucleons (mass number) an isotope of Li mass number talks about 1 atom atomic mass is an average of all atoms

  5. Mass is usually expressed in amu or u 1 amu = 1/12th the mass of a carbon 12 atom 1 amu = 1.66 x 10-27 kg = mass of p+ or n0 an n0 is actually about 5 x 10-4 u more massive an e- has a mass of about 5 x 10-4 u…hmmm

  6. rest energy (E0) = MeV Mega (1 x 106) electron volt (1.60 x 10-19 J) Mass can also be expressed in MeV/c2 1 proton = 938.3 MeV/c2

  7. mass number atomic number (charge)

  8. The electric force between protons is repulsive In a nucleus, there must be a force stronger than the electric force holding the nucleus together

  9. The Strong Force is the force that holds nucleons together. It is independent of charge, it binds neutrons and protons. It only works over very small distances (nuclear).

  10. 10-15 m rnucleus= (1.2x10-15 m )A1/3 where A is the atomic number

  11. Stable “big” atoms have more neutrons than protons, because the electric force acts over greater distances.

  12. Nucleons bound together in a nucleus have less mass (energy) than unbound nucleons binding energy is that energy that holds the nucleus together Ebind = Dmc2 mass defect the difference in mass

  13. Find the total binding energy of Al 27 if it has a mass of 26.981534 u. H 1 has a mass of 1.007825 u and the mass of a n0 is 1.008665 u. There are 931.50 MeV/u. Al is atomic # 13, so it has 13(1.007825 u) of p+ and e- = 13.101725 u. Al has 14 n0 = 14(1.008665 u) = 14.12131 u Dm = (13.101725 + 14.12131) u – 26.981534 u = .241501 u x 931.50 MeV/u = 225 MeV

  14. Nuclear Decay Alpha Decay: an a particle (He nucleus) escapes the strong force by quantum tunneling. The nucleus loses 2 p+ and 2 n0.

  15. Beta decay: the weak force causes an up quark to change into a down quark, or vice versa. This causes a proton to change into a neutron or vice versa. neutron proton electron b- antineutrino

  16. proton neutron positron b+ neutrino A thallium 208 nucleus emits a b- particle (an electron). Write the equation.

  17. Gamma Decay: a nucleon can be in an excited state, needing to get rid of energy to reach a stable energy level. The nucleon emits a photon, but the structure of the nucleus stays the same.

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