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Atomic physics (cont.) - Tran Thi Ngoc Dung

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  1. ATOMIC PHYSICS (CONT.) z Parametric Curve 2 z l 2 = 2 q Y20 x20( t) |Y20|0 y x Y  3cos2 q 1 2,0 2 Parametric Curve 1.2 2 2 0 2 z 1 2 y20( t) 2 q x21(t) |Y210| Re{Y } 21 y Y  sin q cosq 2,1 1.2 1 x Parametric Curve 1.2 1 0 1 1 1.2 y21(t) 1.2 q z x(t) 0 - |Y22| + 2 Re{Y22} y Y2,2  sin q 1.2 1 + 1 0 1 x - 1.2 y(t) 1.2 Tran Thi Ngoc Dung – Huynh Quang Linh – Physics A2 HCMUT 2016
  2. OUTLINE • Orbital angular momentum • Orbital magnetic momentum. • Spin angular momentum • Spin magnetic momentum • Zeeman Effect • Finestructure of spectral line • Pauli exclusion principle • The Periodic Table • Bosons and Fermions
  3. Orbital Angular Momentum Motion of electron around the nucleus: orbital motion Intrinsic motion of electron: spin motion (rotation + around its axis) v Orbital Angular Momentum L 1) The vector does not point in one specific direction. 2) The magnitude of Orbital Angular Momentum is quantized: L  ( 1)  0,1,2, ,n 1 3) The projection of vector on a direction is quantized: ℓ orbital quantum number L m m 0, 1, 2, ,  m : orbital magnetic quantum z number
  4. Orbital Angular Momentum L  ( 1) Lz m Example:for 2, m 0, 1, 2 L  6 Lz 0, , 2 z L  6 m=2 2  q m=1 m=0 O  m=-1 2 L m m=-2 cosq z  L ( 1)
  5. Orbital Magnetic Dipole Moments Consider electron moving with velocity v e z on a circular Bohr orbit of r. The orbital L angular momentum is 2 L mevrez mer ez + m , e Electron, having a charge –e, in the e_ motion around the nucleus, produces a  v currrent i: e e i  i T 2 Current loop produces a magnetic field, with a magnetic moment e er2  iS r2e e 2 z 2 z Vectors L and are in opposite direction  e L 2me
  6. Orbital Magnetic Dipole Moment  e e  B ( 1)  L  m L 2me 2me z B for  2 m 0; 1; 2 The magnitude of Orbital  B 6 Magnetic Momen is   0,   ,  2 quantized z B B e  ( 1) B ( 1) 2me z  B 6 m=-2  2B The projection of vector  on a direction is quantized: B m=-1 m=0 e e e  L m  m m O z z   B  2me 2me 2me B m=1 2B m=2 e 24 2  9.274 10 A.m Bohr Magneton B 2m
  7. Spin angular momentum In the spin motion, electron has spin angular momentum S Electron has the elctric charge –e, in the spin motion, it produces a current i. the current produces magnetic field. The current loop has spin magnetic momentum s Vectors S and sare in opposite direction  e s S me + S _ s
  8. Spin angular momentum The magnitude of Spin Angular Momentum S is quantized 3 S  s(s 1)  2 s 1/ 2 The projection of vector S 3 S  z on a direction is quantized: 2  ms=1/2 Sz ms  / 2 2 ms 1/ 2 O  ms spin magnetic quantum number 2 ms=-1/2
  9. SPIN MAGNETIC MOMENT s e e s S S me me e e 3 e s B 3 s S  3 B 3 me me 2 2me sz B e e 1 sz Sz  B me me 2 1 1 s , m 2 s 2  3 z 3 s B S  z 2 ms=-1/2  B  ms=1/2 2 O O  B ms =1/2 2 ms=-1/2
  10. Zeeman Effect Zeeman effect is the effect of splitting a spectral line into 3 B 0 lines in the presence of a static magnetic field. E2  Electron has orbital magnetic moment , in the external magnetic field B , it has an additional E1 potential energy: E .B zB mB.B ' Energy level E1 becomes : E1 E1 E1 E1 m1B.B ' Energy level E2 becomes : E2 E2 E2 E2 m2B.B ' ' ' Transition from E2 ' to E1 : E2 E1 E2 E1 (m2 m1)B.B energy of photon : h' h mB.B Selection Rules : m 0, 1 BB  m 1 h '  m 0  B  B m 1 h
  11. Example: The transition from 3P to 2S in the presence of magnetic field. Energy Level 2S : (ℓ =0 => m=0) does not split Energy Level 3P : (ℓ =1 => m=0, 1) splits into 3 levels B 0 m = 1 E B 0  B 2 B m=0 3P , ℓ =1 , m=0, 1 m = -1 E 1 2S , ℓ =0 m=0  m 1 m 0 m 1  B  B  B   B h h
  12. Example 2: The transition from 3D to 2P in the presence of magnetic field. Energy Level 2P : (ℓ =1 => m=0, 1) splits into 3 levels Energy Level 3D : (ℓ =2 => m=0, 1, 2) splits into 5levels B 0 m = 2  B B 0 B m = 1 m = 0 E2 3D m = -1 m = -2 E 2P m = +1 1 m = 0  m = -1 m 1 m 0 m 1  B  B  B   B h h
  13. 1 j |  | 2 The energy level is labeled 2 n X j
  14. Energy Levels of Electron ℓ 0 1 2 3 X S P D F j=|ℓ 1/2| 1/2 1/2 3/2 3/2 5/2 5/2 7/2 2 n Xj nS1/2 nP1/2 nP3/2 nD3/2 nD5/2 nF5/2 nF7/2 Energy level splits into 2 levels , except the energy level S Selection Rules  1 , j 0, 1
  15. Example. Examine the transition 2P-3S when we consider the spin of the electron 3S 2 3 S1/ 2 1 2 22P 2P 3/ 2 22 P j 0 j 1 1/ 2  1  1 Without When we consider the spin considering of the electron, the spectral the spin of line is a double spectral one the electron  22 P 32S 1 1/ 2 1/ 2 2 2 2 2 P3/ 2 3 S1/ 2
  16. Example. Examine the transition of 2P-3D when we consider the spin of the electron 2 3D 3 D5/ 2 32 D  3/ 2 1 2 3 2 2P 2 P3/ 2 22 P j 0 j 1 j 1 1/ 2  1  1  1 Without Considering spin considering spin 3 spectral lines: Triple spectral 1 spectral line 2 2 1 2 P3/ 2 3 D3/ 2 2 2 2 2 P1/ 2 3 D3/ 2 2 2 3 2 P3/ 2 3 D5/ 2
  17. The Periodic Table 1. The Pauli’s exclusion principle states that no two electrons can occupy the same quantum-mechanical state in a given system. That is, no two electrons in an atom can have the same values of all four quantum numbers n, ℓ, m, ms . 2. For a given principal quantum number n, there are 2n2 quantum states. n l m ms Maximum Maximum number of number of electrons electrons in the in the shell subshell n=1 l=0 m=0 ms= 1/2 2 2 n=2 l=0 m=0 m = 1/2 2 s 8 l=1 m=0, 1 ms= 1/2 6 n=3 l=0 m=0 ms= 1/2 2 l=1 m=0, 1 ms= 1/2 6 18 l=2 m=0, 1, 2 ms= 1/2 10
  18. Classification of particles Particle Spin Example Obey Pauli Exclusion principle Fermion Half Interger electron, proton, Yes ½, 3/2 nơtron Boson 0 or Interger Photon, Mezon , No 0,1,2 Mezon K
  19.  3, m 0, 1, 2; 3 L  ( 1)  3(3 1)  12 Lz m 0, , 2; 3 L cosq z 0 q 90o L L  cosq z 0.289 q 73.2o L  12 L 2 cosq z 0.577 q 54.74o L  12 L 3 cosq z 0.866 q 30o L  12
  20. a)  E2 E1 (3.4) ( 13.6) 10.2eV 1.24 1.24 (m) 0.12m (eV) 10.2(eV) c 3 108  2.5 1015  0.12 10 6   a) E  .B m B   z B 6 24 '    BB 0.12 10 9.274 10 2.2 12 E2 E2 m2BB  1.478 10 m '   h 2.5 1015 6.626 10 34 E1 E1 m1BB Increase h' h (m2 m1)BB '  BB  B '  B h  B  '  B h
  21. Electron Spin and Nuclear Spin  So, we need FOUR quantum numbers to specify the electronic state of a hydrogen atom 1 1  n, l, ml, ms (where ms = - /2 and + /2)  Actually, the proton in the H atom ALSO has a spin, which is described by an additional quantum number, mp  The energy difference between the two proton spin states in a magnetic field is 660 times smaller than for electron spin states!  But There are many more unpaired proton spins than unpaired electron spins in ordinary matter. Our bodies have many unpaired protons in H2O. Detect them In order to image tissue of various types, Magnetic Resonance Imaging detects the small difference in the numbers of “up” and “down” hydrogen proton spins generated when the object studied is placed in a magnetic field. Nobel Prize (2003): www.beckman.uiuc.edu/research/mri.html Lauterbur (UIUC)
  22. Nuclear Spin and MRI: Example  Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are only two possible spin directions with definite energy. The energy difference between these states is E=2pB, -26 with p = 1.41 x 10 J /Tesla. B 0 B=0 E 2pB B Question 1: The person to be scanned by an MRI machine is placed in a strong magnetic field, with B=1 T being a typical value. What is the energy difference between spin-up and spin-down proton states in this field? Question 2: What is the frequency f of photons that can be absorbed by this energy difference?
  23. Nuclear Spin and MRI: Example  Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is E=2pB, -26 with p = 1.41 x 10 J/Tesla. B 0 B=0 E 2p|B| B Question 1: The person to be scanned by an MRI machine is placed in a strong magnetic field, with B=1 T being a typical value. What is the energy difference between spin up and down proton states in this field? Solution: -26 E 2pB = 2 x (1.41 x 10 J/T) x 1 T = 2.82 x 10-26 J = 2.82 x 10-26 J x 1 eV/ 1.6 x 10-19 J = 1.76 x 10-7 eV
  24. Nuclear Spin and MRI: Example  Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is E=2pB, -26 with p = 1.41 x 10 J/Tesla. B 0 B=0 E 2pB B Question 2: What is the frequency f of photons that can be absorbed by this energy difference?
  25. Nuclear Spin and MRI: Example  Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies. The nucleus is a proton with spin ½, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is E=2pB, -26 with p = 1.41 x 10 J/Tesla. B=0 B 0 E 2p|B| B Question 2: What is the frequency f of photons that can be absorbed by this energy difference? Solution: Energy conservation: Ephoton = hf = E so: f = E / h = 2.82 x 10-26 J / 6.626 x 10-34 J-sec = 42 MHz Radio waves