Complex atoms Structure of nucleus PHY123 11/24/2018 Lecture XVIII

Concepts Quantum numbers, quantum state Pauli principle Periodic table Isotopes Binding energy Fission and fusion 11/24/2018 Lecture XVIII

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) 11/24/2018 Lecture XVIII

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-state is possible 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

Possible number of states 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 given n  2n2 possible 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 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

Periodic table 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 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 11/24/2018 Lecture XVIII

Electron structure 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 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 11/24/2018 Lecture XVIII

Test problem #1 The following configuration is allowed 1s22s22p63s3 True False 11/24/2018 Lecture XVIII

Test problem #2 The following configuration is allowed 1s22s22p63s23p54s2 True False 11/24/2018 Lecture XVIII

Test problem #3 The following configuration is allowed 1s22s22p62d1 True False 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

Nucleus radius Volume proportional to the number of nucleons: 11/24/2018 Lecture XVIII

Isotopes Mass of the nucleus does not determine chemical properties Atomic 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) 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

Binding energy per nucleon Fusion Fission Most tightly woven nuclei in the middle of the Mendeleev’s table Energetically most favorable place Too many protons Electrostatic repulsion Not enough nucleons to build up strong force 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII

Chain reaction Uncontrollable chain reaction – fission bomb Controllable chain reaction – nuclear reactors 11/24/2018 Lecture XVIII

Nuclear reactor 11/24/2018 Lecture XVIII

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 11/24/2018 Lecture XVIII