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. Content 3-Dimensional Potential Well Product Wavefunctions Concept of degeneracy Early Models of the Hydrogen Atom Planetary Model Quantum Modifications Schrödinger’s Equation for the Hydrogen Atom Ground state solution Spherically-symmetric excited states (“s-states”)
3. Particle in a 3D Box (1) The extension of the Schrödinger 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 Schrödinger 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
4. 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
5. 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
6. Energy levels (1) Now back to a 3D cubic box: z 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
7. Energy levels (2) Now consider a non-cubic box: z 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.
8. 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.
9. 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.
10. 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 1515 a 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
11. 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 1515a0 0 r 20a0 radial wave functions radial probability densities, P(r)
12. 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: a. a > a 22 0,He 0 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
13. 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.
14. 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 %
15. 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
16. 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.
17. 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 eV·nm 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.
18. 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 · 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  r mr circular motion 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.
19. Appendix: Solving the SEQ for H deriving ao and E 2 − 2 1 2 e Substituting R ( r ) = Ne − a r 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)