Physics 2 - Lecture 10: Atomic Physics - Huynh Quang Linh

Nội dung text: Physics 2 - Lecture 10: Atomic Physics - Huynh Quang Linh

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. 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
3. 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
4. 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
5. 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
6. 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
7. 1 j |  | 2 The energy level is labeled 2 n X j
8. 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
9. 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
10. 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
11. 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
12. 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
13. 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