Lecture 18: 3D Review, Examples

Transcription

1 Lecture 18: 3D Review, Examples A real (2D) quantum dot Lecture 18, p 1

2 Lect. 16: Particle in a 3D Box (3) The energy eigenstates and energy values in a 3D cubical box are: nxπ nyπ nzπ ψ = N sin x sin y sin z L L L E 2 h = + + 8mL ( n n n ) n n n 2 x y z x y z where n x,n y, and n z can each have values 1,2,3,. y z L L L x This problem illustrates two important points: Three quantum numbers (n x,n y,n z ) are needed to identify the state of this three-dimensional system. That is true for every 3D system. More than one state can have the same energy: Degeneracy. Degeneracy reflects an underlying symmetry in the problem. 3 equivalent directions, because it s a cube, not a rectangle. Lecture 18, p 2

3 Consider a particle in a 2D well, with L x = L y = L. 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 (2,2) < E (1,3) = E (3,1) Act 1 2. If we squeeze the box in the x-direction (i.e., L x < L y ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) Lecture 18, p 3

4 Solution Consider a particle in a 2D well, with L x = L y = L. 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 (2,2) < E (1,3) = E (3,1) E (1,3) = E (3,1) = E ( ) = 1 E E (2,2) = E ( ) = 8 E E 2 h 8mL 2 2. If we squeeze the box in the x-direction (i.e., L x < L y ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) Lecture 18, p 4

5 Solution Consider a particle in a 2D well, with L x = L y = L. 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 (2,2) < E (1,3) = E (3,1) E (1,3) = E (3,1) = E ( ) = 1 E E (2,2) = E ( ) = 8 E E 2 h 8mL 2 2. If we squeeze the box in the x-direction (i.e., L x < L y ) compare E (1,3) with E (3,1). a. E (1,3) < E (3,1) b. E (1,3) = E (3,1) c. E (1,3) > E (3,1) Because L x < L y, for a given n, E for x motion is larger than E for y motion. The effect is larger for larger n. Therefore, E (3,1) > E (1,3). Example: L x = ½, L y = 1: E (1,3) = 13 E (3,1) = 37 We say the degeneracy has been lifted. Lecture 18, p 5

6 Non-cubic Box Consider a non-cubic box: z The box is stretched along the y-direction. What will happen to the energy levels? Define E o = h 2 /8mL 1 2 L 1 x E y L 1 L 2 > L 1 11E o 9E o 6E o 3E o Lecture 18, p 6

7 Consider a non-cubic box: Solution z The box is stretched along the y-direction. What will happen to the energy levels? Define E o = h 2 /8mL 1 2 L 1 x E h E = E n + n + n 2 ( 2 2 ) ( 2 ) nxnynz x z 2 y 8mL2 Smaller than E y L 1 L 2 > L 1 11E o 9E o 6E o 3E o The others are left for you. (2,1,1) (1,1,2) D=2 (1,2,1) D=1 (1,1,1) D=1 1: The symmetry is broken for y, so the 3-fold degeneracy is lowered. A 2-fold degeneracy remains, because x and z are still symmetric. 2: There is an overall lowering of energies due to decreased confinement along y. Lecture 18, p 7

8 Radial Eigenstates of Hydrogen Here are graphs of the s-state wave functions, R no (r), for the electron in the Coulomb potential of the proton. The zeros in the subscripts are a reminder that these are states with l = (zero angular momentum!). E 5 R 1 R 2 R ev R r e r r ( ) r / a 1, 2 ħ = mκe 4a R r a 2.53 nm r / 2a 2,( ) 1 e 2 a r r The Bohr radius of the H atom. 1a R r 2r r 15a r / 3a 3,( ) e a 3 a r ev E 2 4 mκ e = 13.6 ev 2 2ħ The ground state energy of the hydrogen atom. One factor of e or e 2 comes from the proton charge, and one from the electron ev Lecture 18, p 8

9 Wave Function Normalization What is the normalization constant for the hydrogen atom ground state? ψ θ φ r / ao 1( r,, ) = NR1( r ) = Ne 5 R 1 r 4a Lecture 18, p 9

10 Solution What is the normalization constant for the hydrogen atom ground state? ψ θ φ r / ao 1( r,, ) = NR1( r ) = Ne 5 R 1 The probability density is ψ 2 = N 2 exp(-2r/a o ). In 3D, this means probability per unit volume. r 4a We require that the total probability = 1: ψ 2 dv = 1 dv = r 2 sinθ dr dθ dφ With spherical symmetry, the angular integrals give 4π, so we are left with: / 4πN r e r a dr 1 ψ r = 1 e r / ao 1( ) 3 πao = N = 1 πa 3 o You can look it up! Normalized ground-state wave function of hydrogen Lecture 18, p 1

11 Probability Calculation Estimate the probability of finding the electron within a small sphere of radius r s =.2 a o at the origin. ψ(r) = R 1 (r) r s 5 4a ψ = / o ( r ) Ne r a Lecture 18, p 11

12 Solution Estimate the probability of finding the electron within a small sphere of radius r s =.2 a o at the origin. If it says estimate, don t integrate. The wave function is nearly constant near r = : 1 1 o ψ () = e = πa / a 3 3 o πao r s 5 ψ = ψ(r) = R 1 (r) / o ( r ) Ne r a 4a Simply multiply ψ 2 by the volume V = (4/3)πr s3 : 2 4 r s Probability = ψ () V =.1 3 ao 3 Lecture 18, p 12

13 Maximum Radial Probability At what radius are you most likely to find the electron? ψ = / o ( r ) Ne r a 5 r r max 4a Lecture 18, p 13

14 Solution At what radius are you most likely to find the electron? Looks like a no-brainer. r =, of course! 5 ψ = / o ( r ) Ne r a Well, that s not the answer. r You must find the probability P(r) r that the electron is in a shell of thickness r at radius r. For a given r the volume, V, of the shell increases with radius. The radial probability has an extra factor of r 2 : r max r max 4a r / a o P( r ) r = ψ ( r ) V = Cr e r Set dp/dr = to find: r max = a! 5 P(r) r 2 ψ 2 More volume at larger r. V = 4πr 2 r r r r 4a No volume at r =. Lecture 18, p 14

15 Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the effective Bohr radius a,he with the usual Bohr radius for hydrogen, a : a. a,he > a b. a,he = a c. a,he < a Act 2 r 2 ħ mκe = a 2.53 nm The Bohr radius of the H atom. 2. What is the ratio of ground state energies E,He /E,H? a. E,He /E,H = 1 b. E,He /E,H = 2 c. E,He /E,H = 4 Lecture 18, p 15

16 Consider an electron around a nucleus that has two protons, like an ionized Helium atom. 1. Compare the effective Bohr radius a,he with the usual Bohr radius for hydrogen, a : a. a,he > a b. a,he = a c. a,he < a 2. What is the ratio of ground state energies E,He /E,H? a. E,He /E,H = 1 b. E,He /E,H = 2 c. E,He /E,H = 4 Solution Look at how a depends on the charge: 2 2 ħ ħ a a a 2, He = mκe mκ(2 ee ) 2 This should make sense: more charge stronger attraction electron sits closer to the nucleus Lecture 18, p 16

17 Consider an electron around a nucleus that has two protons, (an ionized Helium atom). 1. Compare the effective Bohr radius a,he with the usual Bohr radius for hydrogen, a : a. a,he > a b. a,he = a c. a,he < a 2. What is the ratio of ground state energies E,He /E,H? a. E,He /E,H = 1 b. E,He /E,H = 2 c. E,He /E,H = 4 Solution Look at how a depends on the charge: 2 2 ħ ħ a a a 2, He = mκe mκ(2 ee ) 2 This should make sense: more charge stronger attraction electron sits closer to the nucleus Clearly the electron will be more tightly bound, so E,He > E,H. How much more tightly? Look at E : mκ e mκ (2 e) e E = E = = 4E 2ħ 2ħ, H 2, He 2, H In general, for a hydrogenic atom (only one electron) with Z protons: E = Z E 2, Z, H Lecture 18, p 17

18 Stern-Gerlach Experiment & Electron Spin In 1922, Stern and Gerlach shot a beam of Ag atoms (with l = ) through a non-uniform magnetic field and detected them at a screen. oven B??? screen We can think of the atoms as tiny magnets (they have a magnetic moment) being directed through the field. They are randomly oriented: S N S N N N S S N S Strong B Weak B Lecture 18, p 18

19 Act 3 1. Consider a magnet in an inhomogeneous field, as shown. Which way will the magnet feel a force? a. Up b. Down c. Left d. Right e. No force N S Stronger B Weaker B 2. The magnets (i.e., atoms) leave the oven with random orientations. What pattern do you expect on the screen? Lecture 18, p 19

20 Solution 1. Consider a magnet in an inhomogeneous field, as shown. Which way will the magnet feel a force? N S Stronger B a. Up b. Down c. Left d. Right e. No force The N pole is in a stronger field than the S pole, so its upward force dominates. N S N S Weaker B 2. The magnets (i.e., atoms) leave the oven with random orientations. What pattern do you expect on the screen? Lecture 18, p 2

21 Solution 1. Consider a magnet in an inhomogeneous field, as shown. Which way will the magnet feel a force? Stronger B a. Up b. Down c. Left d. Right e. No force The N pole is in a stronger field than the S pole, so its upward force dominates. N S Weaker B 2. The magnets (i.e., atoms) leave the oven with random orientations. What pattern do you expect on the screen? oven B We expect a blob, because the position depends on the random rotation angle. Lecture 18, p 21

22 B-field off: No splitting B-field on: 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. Lecture 18, p 22

23 Back to the Stern-Gerlach Experiment oven B screen You will analyze this experiment in discussion. The beam split in two! This marked the discovery of a new type of angular momentum, with an m s quantum number that can take on only two values: (s = ½) m s = ±½ The new kind of angular momentum is called the electron SPIN. Why? If the electron were spinning on its axis, it would have angular momentum and a magnetic moment (because it s charged) regardless of its spatial motion. However, this spinning ball picture is not realistic, because it would require the point-like electron to spin 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. Note: Most particles have spin (protons, neutrons, quarks, photons ) Lecture 18, p 23

24 Electron Spin We need FOUR quantum numbers to specify the electronic state of a hydrogen atom. n, l, m l, m s (where m s = -½ and +½) Actually, the nucleus (a proton) also has spin, so we must specify its m s as well We ll work some example problems next time. Lecture 18, p 24

25 Electron Magnetic Moment Because the electron has a charge and angular momentum, it has a magnetic moment, with magnitude: µ e = J/T. One consequence of the quantization of angular momentum is that we only ever measure the spin (and hence the magnetic moment) to be pointing up or down (where the axis is defined by any applied magnetic field). [Note: Because the charge of the electron is negative, the spin and magnetic moment point in opposite directions!] In a uniform magnetic field (B = B z z), a magnetic moment has an energy (Phys. 212): E = -µ. B = -µ z B z Thus, for an electron, the two spin states have two energies: B=: B : B +µ e B -µ e B E = 2µ e B Note: These arrows represent magnetic moment, not spin... Lecture 18, p 25

26 FYI: The real value of µ e There are relatively simple arguments that predict µ e = µ B eћ/2m = x 1-24 J/T In reality, the measured mag. moment of the electron is a bit bigger: µ e = x 1-24 J/T The effect is small: µ e /µ B = (42) [Yes, it has been measured that well in fact, it s one of the most precisely known quantities today.] What causes the discrepancy? It comes from the fact that: Magnetic (and electric) effects essentially arise from the exchange of virtual photons. Sometimes these can, for a very short time, become an electron-positron pair (which then annihilate each other). There are lots of other exotic processes too. When all these are taken into account, our current best theoretical prediction for the value of µ e /µ B = (27) This is agreement to at least 12 decimal places!!