Physics A2 - Lecture 10: Particles in 3D Potentials and the Hydrogen Atom - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 10: Particles in 3D Potentials and the Hydrogen Atom - Huynh Quang Linh

1. “Anyone who can contemplate quantum mechanics without getting dizzy hasn’t understood it.” Neils Bohr Discussion: Is there a particle flying faster than light speed ?
2. Lecture 10: Particles in 3D Potentials and the Hydrogen Atom 1 r / ao  (x, y, z) (x) (y) (z)  ( r ) 3 e ao z 11 1 r = a0 L P(r) L g(x) 0.5 x 00 0 L 0 0 r2 4a0 4 2 0 x 4 h 2 2 2 13.6 eV En n n  nx ny nz En x y z 8mL2 n2
3. Content  3-Dimensional Potential Well  Product Wavefunctions  Concept of degeneracy  Early Models of the Hydrogen Atom  Planetary Model  Quantum Modifications  Schrửdinger’s Equation for the Hydrogen Atom  Ground state solution  Spherically-symmetric excited states (“s-states”)
4. Quantum Particles in 3D Potentials A real (3D)  So far, we have considered quantum particles bound in one-dimensional potentials. This “quantum dot” situation can be applicable to certain physical systems but it lacks some of the features of many “real” 3D quantum systems, such as atoms and artificial quantum structures: (www.kfa-juelich.de/isi/)  One consequence of confining a quantum particle in two or three dimensions is “degeneracy” the occurrence of several quantum states at the same energy level.  To illustrate this important point in a simple system, we extend our favorite potential the infinite square well to three dimensions.
5. Particle in a 3D Box (1)  The extension of the Schrửdinger Equation (SEQ) to 3D is straightforward in cartesian (x,y,z) coordinates: 2 d 2 d 2 d 2 U( x,y,z ) E   (x, y, z) 2 2 2 where 2m dx dy dz 1 2 2 2 Kinetic energy term in the like px py pz 2m Schrửdinger Equation  Let’s solve this SEQ for the particle in a 3D box: z outside box, x or y or z L L L x This simple U(x,y,z) can be “separated” y U(x,y,z) = U(x) + U(y) + U(z) L
6. Particle in a 3D Box (2)  So the Schrửdinger Equation becomes: 2 d 2 d 2 d 2 U( x ) U( y ) U( z )  E 2 2 2 2m dx dy dz and the wavefunction    ( x , y , z ) can be “separated” into the product of three functions:  (x, y, z) (x) (y) (z)  So, the whole problem simplifies into three one-dimensional equations that we’ve already solved in Lecture 7. 2 2 2 2  d (x) nx h nx U(x) (x) E (x) n (x) N sin x Enx  2m dx2 x L 2m 2L 2 2 2 2  d (y) ny h n (y) N sin y E y 2 U(y) (y) Ey (y) n ny 2m dy L 2m 2L Likewise for (z) graphic
7. Particle in a 3D Box (3)  So, finally, the eigenstates and associated energies for a particle in a 3D box are: z nx ny nz  N sin x sin y sin z L L L L 2 x h 2 2 2 En n n  nx ny nz x y z 8mL2 L y where nx,ny, and nz can each have values 1,2,3, . L  This problem illustrates 2 important new points. (1) Three ‘quantum numbers’ (nx,ny,nz) are needed to completely identify the state of this three-dimensional system. (2) More than one state can have the same energy: “Degeneracy”. Degeneracy reflects an underlying symmetry in U(x,y,z) 3 equivalent directions
8. Lecture 10, exercise 1 Consider a particle in a two-dimensional (infinite) well, with Lx = Ly. 1. Compare the energies of the (2,2), (1,3), and (3,1) states? a. E(2,2) > E(1,3) = E(3,1) b. E(2,2) = E(1,3) = E(3,1) c. E(1,3) = E(3,1) > E(2,2) 2. If we squeeze the box in the x-direction (i.e., Lx E(3,1)
9. Lecture 10, exercise 1 Consider a particle in a two-dimensional (infinite) well, with Lx = Ly. 1. Compare the energies of the (2,2), (1,3), and (3,1) states? a. E(2,2) > E(1,3) = E(3,1) 2 2 E(1,3) = E(1,3) = E0 (1 + 3 ) = 10 E0 b. E(2,2) = E(1,3) = E(3,1) 2 2 c. E(1,3) = E(3,1) > E(2,2) E(2,2) = E0 (2 + 2 ) = 8 E0 2. If we squeeze the box in the x-direction (i.e., Lx E 2 (1,3) (3,1) h2 n 2 n E  x y nnxy 22 8m Lxy L Example: Lx = Ly/2 h2 E 4 n22 n E  4 9; E  36 1 (nxy n )2 x y (1,3) (3,1) 8mLy
10. Energy levels (1) z  Now back to a 3D cubic box: Show energies and label (nx,ny,nz) for the first L 11 states of the particle in the 3D box, and write the degeneracy D for each allowed energy. x 2 2 Use Eo= h /8mL . L y L E
11. Energy levels (1) z  Now back to a 3D cubic box: Show energies and label (nx,ny,nz) for the first L 11 states of the particle in the 3D box, and write the degeneracy D for each allowed energy. x 2 2 Use Eo= h /8mL . L y L (nx,ny,nz) E 2 h 2 2 2 En n n nx ny nz x y z 8mL2 nx,ny,nz = 1,2,3, 6Eo (2,1,1) (1,2,1) (1,1,2) D=3 3Eo (1,1,1) D=1
12. Energy levels (2) z  Now consider a non-cubic box: Assume that the box is stretched only along L the y-direction. What do you think will 1 happen to the cube’s energy levels below? x h2 h2 E n 2 n 2 n 2 L > L nxnynz 2 x z 2 y y 2 1 8mL1 8mL2 E L (nx,ny,nz) 1 11Eo 9Eo 6Eo 3Eo
13. Energy levels (2) z  Now consider a non-cubic box: Assume that the box is stretched only along L the y-direction. What do you think will 1 happen to the cube’s energy levels below? x 2 2 h 2 2 h 2 En n n nx nz ny x y z 2 2 y L2 > L1 8mL1 8mL2 E L (nx,ny,nz) 1 11E (1) The symmetry of U is o “broken” for y, so the “three- fold” degeneracy is lowered a 9E o ”two-fold” degeneracy remains due to 2 remaining equivalent 6Eo (2,1,1) (1,1,2) D=2 directions, x and z. (1,2,1) D=1 3E (2) There is an overall lowering o (1,1,1) D=1 of energies due to decreased confinement along y.
14. Another 3D System: The Atom -electrons confined in Coulomb field of a nucleus Early hints of the quantum nature of atoms:  Discrete Emission and Absorption spectra Atomic hydrogen  When excited in an electrical discharge, atoms emitted radiation only at discrete wavelengths  Different emission spectra for different atoms l (nm)  Geiger-Marsden (Rutherford) Experiment (1911): Au  Measured angular dependence of a particles (He ions) scattered from gold foil. a  v  Mostly scattering at small angles supported the “plum pudding” model. But  Occasional scatterings at large angles Something massive in there!  Conclusion: Most of atomic mass is concentrated in a small region of the atom a nucleus!
15. Atoms: Classical Planetary Model (An early model of the atom)  Classical picture: negatively charged objects -e (electrons) orbit positively charged nucleus F due to Coulomb force.  There is a BIG PROBLEM with this: +Ze  As the electron moves in its circular orbit, it is ACCELERATING.  As you learned in Physics 212, accelerating charges radiate electromagnetic energy.  Consequently, an electron would continuously lose energy and spiral into the nucleus in about 10-9 sec. The planetary model doesn’t lead to stable atoms.
16. Can the wave nature of the electron eliminate the radiation problem? What if the electron assumed a ‘stationary quantum state’ (like the particle states in our hypothetical boxes)? Since the probability density is static for a stationary state, there would be no radiation of energy. The stability problem would be solved! Is it possible to find a stationary state for the coulomb potential? Before Schrửdinger (but after DeBroglie), you might have tried the following “semi-classical” approach (just as Neils Bohr did, and for which he got the Nobel prize in 1922): p -e -e ? Instead r Imagine this: of this: +Ze +Ze For a ‘standing wave’, the number of wavelengths equals the circumference: nl = 2 r A proper description requires This is valid only for large n the Schrửdinger Equation (Correspondence Principle); see Appendix
17. Potential Energy for the Hydrogen Atom (1)  How do we describe the hydrogen atom quantum mechanically? U(r)  We need to specify U, the potential r energy of the electron: 0 We assume that the Coulomb force between the electron and the nucleus is the force responsible for binding the electron in the atom e2 U( r ) 1 r  9 109 Nm2 / C 2 4 0  We cannot separate this potential energy in xyz coordinates as we did for the 3D box,U(x,y,z) = U(x) + U(y) + U(z) which led to the product wavefunctions, :  x,y,z f x f y f z This spherically symmetric problem must be solved in spherical coordinates.
18. Product Wavefunction in Spherical Coordinates  The solution to the SEQ in spherical coordinates is a product wave function of the form:   (r,,f) R (r)Y (,f) nlm nl lm r z y with quantum numbers: n l and m x f principal orbital magnetic (angular momentum) Rnl is the ‘radial part’, and Ylm are “spherical harmonics.” The Schrửdinger equation in spherical coordinates is complicated. An appreciation of the problem can be gained by considering only spherically symmetric states (wavefunctions with no angular dependence). For these “s-states”, l = 0 and m = 0, and the ‘radial SEQ’ takes the form: 2 1 2 e2 s-state r R ( r ) E R ( r ) Wavefunction: 2 n0 n n0 2m r r r ( r, ,f ) Rn0( r ) KE term PE term
19. Radial Eigenstates of H  The radial eigenstates, Rno(r) , for the electron in the Coulomb potential of the proton are plotted below: “s-state” wavefunctions The zeros in the subscripts below are a reminder that these are states with zero angular momentum. 1 1 1 1 3 2 R10 0.5 R20 R f(x) 0.5 h(x) d4(x) 30 0 0 .5 000 .2 0 0 0 2 4 0 5 10 0 5 10 15a15 0 4a 0 10a0 0 r 0 0 rx 4 0 0 r x 10 0 x 15 2 2r r r / a r r / 2a r /3a0 0 R (r)  1 e 0 R3,0 (r)  3 2 e R1,0(r)  e 2,0 a 3a 2a0 0 0 2 a0  = “Bohr radius” = 0.053 nm me2 e2 1 13.6eV En 2 2 Plug these into radial SEQ (Appendix) 2ao n n
20. Probability Density of Electrons 2 2 Probability density = Probability per unit volume =   Rn0 for s-states. 2 The density of dots plotted below is proportional to Rn0 . “1s state” “2s state” 1 1 1 1 R10 0.5 R20 h(x) f(x) 0.5 0 .2 0 0 00 0 5 10 0 2 4 0 10a0 0 4a 0 r x 10 0 rx 4 0
21. Radial Probability Densities for s-states  Summary of wave functions and radial probability densities for some s-states: 1 1 1 1 P10 R10 Question: If the f(x) 0.5 g(x) 0.5 classical planetary 0 model were 0 0 0 0 0 2 4 00 2 4 0 4a0 4a correct, then 0 rx 4 00 xr 4 0 1 1 .5 what would you R20 P 0.4 20 expect for P(r)? 0.5 h(x) h2(x) 0.2 00 .2 0 0 5 10 0 10a 0 0 5 10 00 x 10 0 0 10a r 0 rx 10 0 3 2 P30 R30 d4(x) 0 .5 0 0 0 5 10 15 0 rx 15a15 0 0 r 20a0 radial wave functions radial probability densities, P(r)
22. Lecture 10, exercise 2 Consider an electron around a nucleus that has two protons (and two neutrons, like an ionized Helium atom). 1. Compare the “effective Bohr radius” a0,He with the usual Bohr radius for hydrogen, a0: a. a0,He > a0 b. a0,He = a0 c. a0,He < a0 2. What is the ratio of ground state energies E0,He/E0,H? a. E0,He/E0,H = 1 b. E0,He/E0,H = 2 c. E0,He/E0,H = 4
23. Lecture 10, exercise 2 Consider an electron around a nucleus that has two protons (and two neutrons, like an ionized Helium atom). 1. Compare the “effective Bohr radius” a0,He with the usual Bohr radius for hydrogen, a0: Look at how a0 depends on the charge: 22 a. a0,He > a0 a0 aa02 0,He  b. a0,He = a0 m e m(2 e ) e 2 This makes sense – more charge c. a0,He |E | . How much more tightly? b. E0,He/E0,H = 2 0,He 0,H Look at E : c. E /E = 4 0 0,He 0,H e2 (2 e ) e EEE0,H 0, He 4 0, H 2aaoo 2( / 2) In general, for a “hydrogenic” E Z 2 E Z2 0,H 13.6 eV atom with Z protons: nZ, nn22
24. Optical Transitions in H Example  An electron, initially excited to the n = 3 r/a 0 0 20 energy level of the hydrogen atom, falls 0 5 10 15 20 to the n = 2 level, emitting a photon in 0 0 the process. What are the energy and E wavelength of the photon emitted? 3 -5 E2 E (eV) U(r) -10 -15-15 E1 Atomic hydrogen l (nm) Answer: 656 nm Discharge tubes
25. Optical Transitions in H Example  An electron, initially excited to the n = 3 r/a 0 0 20 energy level of the hydrogen atom, falls 0 5 10 15 20 to the n = 2 level, emitting a photon in 0 0 the process. What are the energy and E wavelength of the photon emitted? 3 -5 E 13.6 eV 2 E E (eV) n 2 U(r) n -10 Therefore: 11 E 13.6 eV -15-15 E1 nnif 22 nnif 11 Atomic hydrogen EEphoton 32 13.6 eV 1.9eV 94 hc 1240eV nm l 656nm l (nm) Ephoton 1.9eV You will experimentally measure several transitions in Lab.
26. Ground-state wavefunction for H —three standard problems 1) What is the normalization constant N? r / a ( r ) Ne o Standard procedure: 1 1 2 2 a) The probability density is    = N exp(-2r/ao).  r R10 (r) This is the “probability per unit volume”. f(x) 0.5 b) Integrate over volume:Probability =   2 dV = 1 000 2 0 2 4 c) For spherical symmetry, dV = 4 r dr 0 4a0 0 x 4 2 2 2r / a 2 1 4 N r e o dr 1 yields N (integral table) 3 0 ao 1 r / ao Ground-state wavefunction  ( r ) 3 e ao for hydrogen
27. Ground-state wavefunction for H —three standard problems 2) Estimate the probability of finding 1 1 the electron within a small sphere of  r R10 (r) radius rs = 0.2 ao at the origin. f(x) 0.5 rs a) If it says “estimate”, don’t “integrate”. 000 0 2 4 0 4a0 b) The wavefunction is nearly constant near r = 0: 0 x 4 r / a ( r ) Ne o 1 0 / ao 1 (0 ) 3 e 3 ao ao 2 3 c) Simply multiply    by the volume V = (4/3) rs . Probability (0 ) 2 V Answer: 1.07 %
28. Ground-state wavefunction for H —three standard problems 2) Estimate the probability of finding 1 1 the electron within a small sphere of  r R10 (r) radius rs = 0.2 ao at the origin ? f(x) 0.5 rs a) If it says “estimate”, don’t “integrate”. 000 0 2 4 0 4a0 b) The wavefunction is nearly constant near r = 0:0 x 4 r / a ( r ) Ne o 1 0 / ao 1 (0 ) 3 e 3 ao ao 2 3 c) Simply multiply    by the volume V = (4/3) rs . 3 2 4 r s = 0.0107 Probability (0 ) V 3 ao
29. Ground-state wavefunction for H —three standard problems 3) At what radius are you most likely 1 1 r / a to find the electron? ( r ) Ne o Looks like a no-brainer. r = 0, of course! f(x) 0.5 Well, that’s not the answer. You must find the r probability P(r) r that the electron is in a shell 0 00 0 r 2 4 0 ma 4a0 of thickness r at radius r. The volume of the 0 x 4 x shell V increases with radius: rmax = ? 2 Probability P(r) r ( r ) V 1 1 P(r) r 2 -2r/ao C r e r  4 r22 P(r)g(x) 0.5 r Set dP/dr = 0 to find rmax r 0 0 0 2 4 0 r 4a0 0 x 4 2 V = 4 r r rmax = Answer: ao
30. Example Problems (1) a) What is the energy of the second excited state of a 3-D cubic well? b) How many states have this energy? Answer: 9 Eo, 3
31. Example Problems (1) a) What is the energy of the second excited state of a 3-D cubic well? b) How many states have this energy? Solution: Starting with E211 (the first excited state), we must 2 2 2 a) E311 = (3 +1 +1 ) Eo = 11 Eo increase one of the quantum 2 2 2 E221 = (2 +2 +1 ) Eo = 9 Eo numbers. Which choice adds the least energy? E or E ? 2 2 221 311 with Eo= h /8mL There are three distinct ways b) E221 E212 E122 to arrange the three numbers, 2, 2, and 1. The different arrangements correspond to different x, y, and z components of the momentum.
32. Example Problems (2) Consider the three lowest energy states of the hydrogen atom. What wavelengths of light will be emitted when the electron jumps from one state to another? Answer: 122 nm, 102 nm, 653 nm
33. Example Problems (2) Consider the three lowest energy states of the hydrogen atom. What wavelengths of light will be emitted when the electron jumps from one state to another? Solution: E = -13.6 eV/n2, so E = -13.6 eV, E = -3.4 eV, E21 = 10.2 eV 1 2 E = -1.5 eV E = 12.1 eV and 3 . There are three jumps to 31 consider, 2-to-1, 3-to-1, and 3-to-2. The E32 = 1.9 eV photon carries away the energy that the electron loses. l = h/p = hc/E hc = 1240 eVãnm l21 = 122 nm l31 = 102 nm Two wavelengths are all in the ultraviolet. Note that the 3-to-2 transition gives a l32 = 653 nm visible (red) photon, l32 = 653 nm.
34. Example Problems (3) What happens to the ground state radius for the first electron bound to a helium nucleus (Z = 2), i.e., for a helium ion? What is the ratio of its typical radius to that of the H atom? Solution: Ze2 The potential energy for a single electron U Z ( r ) and a nucleus with Z protons involves the r product of those charges, i.e. -e Ze. 2 The Bohr radius, a0, also has the factor a0 mZe 2 e2 for hydrogen; it becomes Ze2 for a nucleus with a single electron. 1 a0 (He) a0 (H ) 2
35. Appendix: Semi-classical model for H atom (Z = 1) 1) For a matter wave in a circular orbit: nl = 2 r l is related to the classical momentum: l = h/p. -e n(h/p = 2 r The product r ã p is h +e n r  p quantized in units of 2  h / 2 r  p n n 1,2,3 (Eq.1) 2) There is also a classical relation between the momentum and radius: 2 e Inverse-r2 force with F 9 r2 Coulomb const.  = 9x10 r p22 m  e (Eq.2) vp22 F ma m Kinetics of circular motion  r mr Classically any orbit radius is possible. Choose r and Equation 2 gives p. The standing wave condition further constrains p and r with Equation 1.
36. Quantum orbits, energy, and angular momentum Quantum (Eq.1) -e condition: r  p n ( n 1,2,3 ) Classical 2 2 kinetics: r  p me for H-atom (Eq.2) +e 2 Solve for r: r n2 n2a Quantized orbits n me2 o 2 2 e 1 13.6eV En ao 2 0.0529 nm 2 2 me 2ao n n “Bohr Radius” Energies of quantized orbits Notice also: Angular momentum L =  r p  = r L r  p n ãp is quantized in units of h-bar. n 1,2,3 In 1913, Neils Bohr proposed angular momentum quantization (L = nħ) to explain the hydrogen spectrum. At the time, DeBroglie waves and quantum mechanics hadn’t been invented. The semi-classical model given above provides correct hydrogenic energies En but fails to explain the H-atom and other atoms in detail. A proper description requires the Schrodinger Equation. ( The picture above is not bad for very large n.)
37. Appendix: Solving the SEQ for H deriving ao and E 2 2 2 ar  1  e  Substituting R ( r ) Ne into r R ( r ) ER ( r ) , we get: 2m r 2 r dr 2 2 1 e 2ae ar a 2re ar e ar Ee ar 2m r r  For this equation to hold for all r, we must have: 2 2 2a  a e 2 AND E m 2m 2 2 me 1 E a 2  2  a0 2ma0  Evaluating the ground state energy: 2 2c2 (197)2 E 13.6eV 2 2 2 6 2 2ma0 2mc a0 2(.51)(10 )(.053)
38. Homework # 6  Can a hydrogen atom absorb a photon whose energy exceeds its binding energy (13.6 eV)?  Find the value of the quantum number for a hydrogen atom that has an orbital radius of 847 pm?  Calculate the binding energy of the hydrogen atom in the first excited state?