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Complex atoms Structure of nucleus

Complex atoms Structure of nucleus. PHY123. Concepts. Quantum numbers, quantum state Pauli principle Periodic table Isotopes Binding energy Fission and fusion. Pauli principle. Electron has spin 1/2 Electron is a fermion:

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Complex atoms Structure of nucleus

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  1. Complex atomsStructure of nucleus PHY123 Lecture XVIII

  2. Concepts • Quantum numbers, quantum state • Pauli principle • Periodic table • Isotopes • Binding energy • Fission and fusion Lecture XVIII

  3. Pauli principle • Electron has spin 1/2 • Electron is a fermion: • Not more than one electron can be in each quantum state (Pauli principle) • Pauli principle is responsible for periodic table (Mendeleev) • NB. If particle has an integer spin (0 or 1) it is a boson - all particles tend to fall in the same state. • Example – photons (lasers use this principle) Lecture XVIII

  4. Electron quantum state • Principle quantum number n=1,2,3,4,… • determines energy level, higher E for higher n • Orbital quantum number l • For each n l can be 0,1,2,3, …(n-1) • l states are leveled by letters • s: l=0; p: l=1; d: l=2; f: l=3; g:l=4 • E.g. n=5, then l can be 0, 1, 2, 3, 4 • Possible l states are s,p,d,f,g • n=1, only l=0 s-stateis possible Lecture XVIII

  5. Electron quantum state z • Orbital quantum number is a vector length l • Its projection on z axis is another q.n. – magnetic quantum number ml • ml can be only integer Lecture XVIII

  6. Electron quantum state z • All electrons have spin=1/2 • It is a vector • Its projection on z axis is another q.n. – spin ms • ms can be only Lecture XVIII

  7. For each given n,l,ml 2 possible states: ms=+1/2; ms=-1/2 For each given n,l  2(2l+1)states: ml=+l, +(l-1), …+2,+1, 0, -1, -2, …-(l-1), -l  (2l+1) For each ml: ms=+1/2; ms=-1/2  2 For each givenn  2n2possible states l=0,1,2,3,4,….(n-1) For each l: ml=+l, +(l-1), …+2,+1, 0, -1, -2, …-(l-1), -l  (2l+1) For each (l,ml) ms=+1/2; ms=-1/2 Possible number of states Lecture XVIII

  8. State labeling • 2p3 means “3 electrons in p state (l=1) at n=2 level” • 4f5 means “5 electrons in f state (l=3) at n=4 level” • States filled up from lower to higher n • For each n from lower to higher l Lecture XVIII

  9. Elements in increasing mass MA=NpMp+NnMn But mass does not determine chemical properties Atomic number Z does. Z=Np-charge of nucleus = number of electrons H Z=1 1 electron n=1, l=0 (s), ml=0 1s1 He Z=2 2 electrons n=1, l=0 (s), ml=0 1s2 2 e: spin up, spin down Li Z=3 3 electrons: 2 e on 1s shell 1 e on n=2, l=0 (s), ml=0 1s22s1 Periodic table Lecture XVIII

  10. n=1,2,3,4…. l=0,1,2,…(n-1) ml=-l,…-1,0,1,…+l ms=+1/2; -1/2 3p4: 4 electrons on n=3, l=1 subshell 2n2 states for each n 2(2l+1) states for each n,l 2 states for each n,l,ml s(l=0), p(l=1), d(l=2), f(l=3) 2 states in each s subshell 6 states in each p subshell 10 states in each d subshell 14 states in each f subshell Electron structure Lecture XVIII

  11. Test problem #1 • The following configuration is allowed 1s22s22p63s3 • True • False Lecture XVIII

  12. Test problem #2 • The following configuration is allowed 1s22s22p63s23p54s2 • True • False Lecture XVIII

  13. Test problem #3 • The following configuration is allowed 1s22s22p62d1 • True • False Lecture XVIII

  14. Atomic spectra • Forbidden (not completely though), allowed transitions (Dl=±1), because photon has spin 1 and angular momentum is conserved • Inner energy shells (called K-shells) “see” almost full potential of the nucleus proportional to (Z-1)2 • Observe atomic spectra in X-ray scattering Lecture XVIII

  15. Nucleons and nuclei • Proton and neutron are called nucleons • 1u=1.66054x10-27kg=931.5MeV/c2 • Protons have positive charge , neutrons have zero charge • A new kind of force – “strong” force – holds nucleons inside nucleus. Unlike EM force strong force is “short distance”  nuclei are very tightly packed Lecture XVIII

  16. Nucleus radius • Volume proportional to the number of nucleons: Lecture XVIII

  17. Isotopes • Mass of the nucleus does not determine chemical propertiesAtomic number Z does. • Same chemical element (Z=Np) can have different number of neutrons –isotopes: • Carbon Z=66 p, but it can have • 5n: 11C, 6n: 12C, 7n: 13C, 8n: 14C, 9n: 15C, 10n: 16C • Only 12C is stable (98.9% of C on Earth) Lecture XVIII

  18. Masses of nuclei and nucleons • Let’s compare nucleus mass with the sum of nucleon masses (from appendix D in Giancoli): • Al: Z=13  13 p, A=27 27-13=14 n • M(13p+14n)=13x1.007276u+14x1.008665=27.215898u • M(Al)=26.981538u • Where did (27.215898-26.981538=0.23436u) go? • Recall that mass is a form of energy • To extract a nucleon from the nucleus you must supply energy W: • M(Al)+W=M(13p)+M(14n) • Hence M(Al)<M(13p)+M(14n) • W is called binding energy Lecture XVIII

  19. Binding energy per nucleon Fusion Fission Too many protons Electrostatic repulsion Most tightly woven nuclei in the middle of the Mendeleev’s table Energetically most favorable place Not enough nucleons to build up strong force Lecture XVIII

  20. Nuclear fission • Discovered by German scientists Otto Hahn and Fritz Strassmann in 1938 • 235U splits into two nuclei when bombarded by neutrons • More neutrons are produced in the reaction: • n+ 235U141Ba+92Kr+3n • This means a chain reaction is possible when the minimum mass of uranium reaches critical mass • Tremendous amount of energy is released Lecture XVIII

  21. Chain reaction Uncontrollable chain reaction – fission bomb Controllable chain reaction – nuclear reactors Lecture XVIII

  22. Nuclear reactor Lecture XVIII

  23. Fusion reactions in the Sun • The key is to bring two protons together close enough for the strong force to overcome electrostatic repulsion – need high T: • 1H+1H2H+e++n • 1H+2H3He+g • 3He+3He4He+1H+1H • This is how heavier (Z>1) elements were produced in the first place (only upto iron or nickel) • Heavier elements are produced in the middle of the starts or supernova explosions by absorbing extra energy • Fission bomb is used to ignite fusion bomb – hydrogen bomb Lecture XVIII

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