Physics A2 - Lecture 11: Angular Momentum, Atomic States, & the Pauli Principle - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 11: Angular Momentum, Atomic States, & the Pauli Principle - Huynh Quang Linh

1. Lecture 11: Angular z Parametric Curve 1.2 1 Momentum, Atomic |Y00| y(t) States, & 0 the Pauliz Principle 1.2 1 1 Parametric0 Curve 1 Parametric Curve 1.2 1.2 1.2 x( t) 1.2 1 |Y | 1 10 |Y11| x10(t) x11(t) 0 0 z 1.2 1 1.2 1 Parametric Curve 1 Parametric0 Curve 1 1 Parametric0 Curve 1 1.2 2 2 1.2 1.2 y10( t) 1.2 1.2 y11( t) 1.2 1 1 |Y20| |Y21| |Y22| x20(t) x21(t) x(t) 0 0 0 2 1.2 1 1.2 1 2 2 0 2 1 0 1 1 0 1 2 y20( t) 2 1.2 y21( t) 1.2 1.2 y(t) 1.2
2. Content  Schrửdinger’s Equation for the Hydrogen Atom  Radial wave functions  Angular wave functions  Angular Momentum 2  Quantization of Lz and L  Spin and the Pauli exclusion principle  Stern-Gerlach experiment  Nuclear spin and MRI
3. Summary of s-states of H-atom  To get the exact eigenstates and energies for the “s-states” of the Coulomb potential, one needs to solve the radial SEQ 2 1 2 e2 r R( r ) ER( r ) 2 2m r r r (spherically  Summary of wave functions for the “s-states”: symmetric)  The zeros in the subscripts below are a reminder that these are states with zero angular momentum. 1 3 1 1 1 R20 2 R30 R10 0.5 d4(x) f(x) 0.5 h(x) 0 0 0 .5 0 .2 0 0 5 10 15 0 2 4 0 5 10 0 15a0 0 0 10a0 0 rx 15 0 0 rx 4a4 0 0 rx 10 2 r 2r r r / 2a0 r /3a0 r / a0 R (r)  1 e R3,0 (r)  3 2 e R1,0(r)  e 2,0 a 3a 2a0 0 0
4. Wavefunction of the H-atom  The solution to the Schrửdinger Equation (SEQ) in spherical coordinates is a product q wave function of the form: r z y nlm(r,q,f) Rnl (r)Ylm(q,f) x f with quantum numbers: n l and m principal orbital magnetic (angular momentum) The Ylm(q,f) are known as “Spherical harmonics”. They are related to the angular momentum of the electron. Last lecture we studied the properties of the radial part. Today we will examine the angular part.
5. Quantized Angular Momentum  The dependence of our wavefunction must have the form Re(y) im r Yelm, (,)qf  Reminder: eimf = cos(mf) + i sin(mf) f  Semi-classical ‘argument’ for angular momentum quantization: An integer number, m, of wavelengths must fit around the circle (radius r) above! 2p r = ml m = 0, 1, 2, 3, m = ‘orbital’ magnetic quantum number Quantization arises because only certain wavelengths fit around path. But, l = h/pt, where pt is the tangential momentum of ‘orbiting’ electron: p r = L = m The angular momentum along a given axis t z can have only quantized values: m = 0, 1, 2, 3, Lz = 0, , 2, 3,
6. The “l” Quantum Number  The quantum number m reflects the component of angular momentum about a given axis. Lz m where m 0, 1, 2,  The quantum number l in the angular wave function Ylm(q,f) tells the total angular momentum L. 2 2 2 2 2 L = Lx + Ly + Lz is quantized. The possible values of L are: L2 l( l 1)2 where l 0,1, 2, Summary of quantum numbers for the H-atom: Principal quantum number: n = 1, 2, 3, . Orbital quantum number: l = 0, 1, 2, , n-1 Orbital magnetic quantum number: m = -l, -(l-1), 0, (l-1), l ( = ml )
7. Classical Picture of L-Quantization e.g., l = 2 L l( l 1)2 2( 2 1)2 6 Lz L = 6  +2 + Classically, the angular momentum 0 vector precesses around the z axis. - L r p -2 Why can’t the orbital angular momentum vector simply point in a specific direction, e.g., along z? If it did, then that would mean that r and p must both be in the x-y plane. But that means that there would be no position uncertainty Dz in the z-direction, nor any momentum uncertainty Dpz, i.e., Dz = Dpz =0, in violation of the Uncertainty Principle.
8. Probability Density of Electrons 2 2 2 Probability density = Probability per unit volume = ynlm  Rnl Ylm 2 The density of dots plotted below is proportional to ynlm . “1s state” “2s state” “2p states” n = 1 2 2 l = 0 0 1 ml = 0 0 0, ±1
9. The Angular Wavefunction, Ylm(q, ) im  The angular wavefunction may be written: Ylm(q, ) = P(q)e where P(q) are polynomial functions of q.  To get some feeling for these angular distributions, we make polar plots of the q-dependent part of |Ylm(q, )| (i.e., P(q)): (Length of the dashed arrow is the magnitude of Ylm as a function of q.) z Parametric Curve Parametricz Curve 1.2 z 1.2 l =0 1 l =1 1 q q Y10 |Y y(t)| |Yx10(t)| 00 0 0 10 0 0 y 1 Y1,0  cosq Y0,0 1.2 1 1.2 1 x 4p 1 0 1 1 0 1 1.2 x( t) 1.2 1.2 Parametricy10( t) Curve 1.2 z 1.2 z 1 Y00 q Re{Y11} |Yx11(t) | q 11 0 0 0 y y x Y1,1 1.21 sin q 1 0 1 x 1.2 y11( t) 1.2
10. The Angular Wavefunction, Ylm(q, ) z Parametricz Curve l = 2 2 2 q Y20 |Yx20(t)| 20 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) |Y21|0 Re{Y21} y Y  sin q cosq x 2,1 1.2 1 Parametric Curve 1.2 1 0 1 1 1.2 y21( t) 1.2 q z x(t) - |Y22|0 + Re{Y } 2 22 y Y2,2  sin q 1.2 1 + 1 0 1 x - 1.2 y(t) 1.2
11. The Angular Wavefunction, Ylm(q, ) Why are these distributions important?  They govern the bonding and chemistry of atoms.  In particular, they determine the angles at which different atoms “bond”. The structure of molecules & solids Historical Labeling of Orbitals Angular momentum quantum # Notation from spectroscopy l = 0 s “sharp” l = 1 p “principle” l = 2 d “diffuse” l = 3 f “fundamental”
12. Effect of l on Radial Wavefunctions Rn,l General properties: n=2 n=3 (1) l < n Total energy must always be larger than rotational part l = 0 l = 0 (2) Number of ‘wiggles’ in Rnl(r) (i.e., radial kinetic energy term) increases with decreasing l l = 1 l = 1 More total energy E in translational than rotation E = Ttrans + Trot + U(r) l = 2 (3) # of zero-crossings = (n – 1) - l
13. Exercise 1  What is the most likely radial position r r / 2a0 of an electron in an n = 2, l = 1 state? R2,1(r)  e a0 (c) r = 5a (a) r = 2a0 (b) r = 4a0 0 Hint: Set the derivative of y2 DV equal to zero and solve for r. (DV for a shell of thickness Dr equals 4pr2Dr.)
14. Solution  What is the most likely radial position r r / 2a0 of an electron in an n = 2, l = 1 state? R2,1(r)  e a0 (c) r = 5a (a) r = 2a0 (b) r = 4a0 0 Hint: Set the derivative of y2 DV equal to zero and solve for r. DV for a 2 2 4 shell of thickness Dr equals 4pr Dr. Therefore, y DV  r exp(-r/ao). d r4 r4e r / a0 e r / a0 4r3 dr a0 Setting this expression equal to zero to find the maximum: r 5 4 0 r 4a 4 P21 a 0 0 h10(x) 2 00 0 0 5 10a10 00 xr 10 0
15. Hydrogen Atom States: Summary  Key Points:  n: principal quantum # 0 5 10 15 r/a0 20  l: orbital quantum # 0  ml: orbital magnetic quantum # n=3 n=2  Energy depends only upon n e2 11 me 24 -5 E e n 22a n2 2 n 2 0 E U(r)  For a given value of n, there -10 are n possible ang. momentum states: l = n-1, n-2, , 2, 1, 0 n=1  For a given angular momentum -15 state l, there are 2l + 1 possible orientations with z- Therefore, a level with quantum axis ml = -l, -(l -1), 0, (l -1), l number n has n2 degenerate states.
16. Hydrogen Atom States: Summary r/a0 0 5 10 15 20 (n,l,ml) 0 n=3 (3,0,0), (3,1,-1), (3,1,0), (3,1,1), (3,2,-2), (3,2,-1), (3,2,0), (3,2,1), (3,2,2) n=2 (2,0,0), (2,1,-1), (2,1,0), (2,1,1) -5 n = 1, 2, 3, E U(r) l = 0, 1, 2, n-1 -10 ml = -l, -(l - 1), 0, (l -1), l n=1 (1,0,0) -15
17. Stern-Gerlach Experiment  In 1922, Stern and Gerlach shot a beam of Ag atoms through a non-uniform magnetic field and detected them at a screen. ??? screen  We can think of the atoms as tiny magnets (they have a magnetic moment) being directed through the field: N Strong B S Weak B
18. Exercise 2 N 1. Consider a magnet in an inhomo- Stronger B geneous field, as shown. Which S way will the magnet feel a force? (Hint: Pretend the field is created N by another magnet above the Weaker B first one and another one below.) S N (a). Up S (b). Down (c). Left (d). Right (e). No force 2. The magnets (i.e., atoms) leave the oven with all orientations; how should the pattern on the screen look?
19. Exercise 2 N 1. Consider a magnet in an inhomo- Stronger B geneous field, as shown. Which S way will the magnet feel a force? (Hint: Pretend the field is created N by another magnet above the Weaker B first one and another one below.) S N (a). Up S (b). Down (c). Left The north pole of the magnet is (d). Right attracted up more than the south (e). No force pole is attracted down. Therefore there is a net upward force. 2. The magnets (i.e., atoms) leave the oven with all orientations; how should the pattern on the screen look? An even distribution, peaked around “no deflection” – some of the magnets are deflected up, some deflected down, some not deflected, etc.
20. Stern-Gerlach Experiment  In 1922, Stern and Gerlach shot a beam of Ag atoms through a non-uniform magnetic field and detected them at a screen. spin You will analyze this experiment in detail in your discussion section. The amazing result: The beam split in two! This marked the discovery of the electron spin. A new type of angular momentum, with a quantum number that can take on only two values: (s = ẵ) ms = ±ẵ
21. B-field off: B-field on: No splitting Two peaks! Gerlach's postcard, dated 8 February 1922, to Niels Bohr. It shows a photograph of the beam splitting, with the message, in translation: “Attached [is] the experimental proof of directional quantization. We congratulate [you] on the confirmation of your theory.”
22. Electron Spin  The new kind of angular momentum is called the electron SPIN. Why call it spin?  If the electron were spinning on its axis, it would have angular momentum and a magnetic moment regardless of its spatial motion.  However, this “spinning” ball picture is not realistic, because it would require that the tiny electron be spinning so fast that parts would travel faster than c !  So we can’t picture the spin in any simple way the electron’s spin is simply another degree-of-freedom available to electron B 0 E mB mzBz B=0 meB DE 2meB| B meB 24 me 9.2848 10 J/Tesla electron magnetic moment 24 mB = e /2me 9.2741 10 J/Tesla (the "Bohr Magneton" related to orbital motion) Note: All particles have spin (e.g., protons, neutrons, quarks, photons)
23. 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): Lauterbur (UIUC) www.beckman.uiuc.edu/research/mri.html
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 ẵ, so in a magnetic field B there are only two possible spin directions with definite energy. The energy difference between these states is DE=2mpB, -26 with mp = 1.41 x 10 J /Tesla. B 0 B=0 DE 2mpB 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?
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 ẵ, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is DE=2mpB, -26 with mp = 1.41 x 10 J/Tesla. B 0 B=0 DE 2mp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 DE 2mpB = 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
26. 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 ẵ, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is DE=2mpB, -26 with mp = 1.41 x 10 J/Tesla. B 0 B=0 DE 2mpB B Question 2: What is the frequency f of photons that can be absorbed by this energy difference?
27. 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 ẵ, so in a magnetic field B there are the only two possible spin directions with definite energy. The energy difference between these states is DE=2mpB, -26 with mp = 1.41 x 10 J/Tesla. B=0 B 0 DE 2mpB| B Question 2: What is the frequency f of photons that can be absorbed by this energy difference? Solution: Energy conservation: Ephoton = hf = DE so: f = DE / h = 2.82 x 10-26 J / 6.626 x 10-34 J-sec = 42 MHz Radio waves
28. Exercise 3 We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a nuclear spin to flip. What therefore must be the angular momentum of each photon? (a). 0 (b). ħ/2 (c). ħ
29. Exercise 3 We just saw that radio-wave photons with energy 1.7 x 10-7 eV can cause a nuclear spin to flip. What therefore must be the angular momentum of each photon? (a). 0 Initial angular mom. = Final ang. (b). ħ/2 momentum (c). ħ electron photon electron Sz () + Sz = Sz () photon electron electron Sz = Sz () - Sz () = ħ/2 – (- ħ/2) = ħ Conclusion – Photons carry angular momentum. In particular, they can carry +ħ, -ħ, or 0, corresponding to right-circular polarization, left-circular polarization, or linear polarization. Particles with an intrinsic spin of nħ are called “bosons”.
30. Recent Breakthrough – Detection of a single electron spin!  (Nature July 14, 2004) IBM scientists achieved a breakthrough in nanoscale magnetic resonance imaging (MRI) by directly detecting the faint magnetic signal from a single electron buried inside a solid sample. Raffi Budakian (started at UIUC Fall ’05) Next step – detection of single nuclear spin (660x smaller).
31. Pauli Exclusion Principle We now want to start building more complicated atoms to study the Periodic Table. For atoms with many electrons (e.g., carbon: 6, iron: 26, etc.) - what energies do they have? From spectra of complex atoms, Wolfgang Pauli (1925) deduced a new rule: “Pauli Exclusion Principle” “In a given atom, no two electrons* can be in the same quantum state, i.e. they cannot have the same set of quantum numbers n, l, ml , ms” I.e., every “atomic orbital with n,l,ml” can hold 2 electrons: ()  Therefore, electrons do not pile up in the lowest energy state, i.e, the (1,0,0) orbital.  They are distributed among the higher energy levels according to the Pauli Principle.  Particles that obey the Pauli Principle are called “fermions” *Note: More generally, no two identical fermions (any particle with spin of ħ/2, 3ħ/2, etc.) can be in the same quantum state.