Physics A2 - Lecture 9: Barrier Penetration and Tunneling - Huynh Quang Linh

Nội dung text: Physics A2 - Lecture 9: Barrier Penetration and Tunneling - Huynh Quang Linh

1. “All of modern physics is governed by that magnificent and thoroughly confusing discipline called quantum mechanics It has survived all tests and there is no reason to believe that there is any flaw in it .We all know how to use it and how to apply it to problems; and so we have learned to live with the fact that nobody can understand it.” Murray Gell-Mann
2. Content How quantum particles tunnel Nuclear Decay Solar Fusion NH3 Maser
3. Application: Tunneling Microscopy Due to the quantum effect of “barrier x Metal penetration,” the electron density of a tip material extends beyond its surface: One can exploit this material STM tip to measure the electron density on a ~ 1 nm material’s surface: material STM tip Real STM tip Na atoms DNA Double on metal: Helix: STM images www-aix.gsi.de/~bio
4. Tunneling Through a Barrier (2) U(x) In general the tunneling coefficient T U0 can be quite complicated (due to the E contribution of amplitudes “reflected” off the far side of the barrier). 0 L x However, in many situations, the barrier width L is much larger than the ‘decay length’ 1/K of the penetrating wave; in this case (KL >> 1) the tunneling coefficient simplifies to: EE 2m −2KL where T Ge G =−16 1 K = 2 (U0 − E) UU00  This is nearly the same result as in the “leaky particle” example! Except for G: • G slightly modifies the 4 3 transmission probability G 2 • G arises from the fact 1 that the amplitude at 0 x = 0 is not a maximum 0 0.25 0.5 0.75 1 E/U0
5. Example: Barrier Tunneling in an STM Let’s consider a simple problem: U(x) An electron with a total energy of E=6 eV U0 approaches a potential barrier with a E height of Uo = 12 eV. If the width of the metal STM tip barrier is L=0.18 nm, what is the 0 L x probability that the electron will tunnel air through the barrier? gap
6. Lecture 9, exercise 1 Consider a particle tunneling through a barrier. U(x) 1. Which of the following will increase the U0 likelihood of tunneling? E a. decrease the height of the barrier b. decrease the width of the barrier 0 L x c. decrease the mass of the particle 2. What is the energy of the particles that have successfully “escaped”? a. initial energy
7. Example: Al wire contacts “Everyday” problem: U(x) You’re putting the electrical wiring in your new U0 house, and you’re considering using Aluminum E outlet wiring, which is cheap and a good conductor. Al wire However, you also know that aluminum tends to contact form an oxide surface layer (Al2O3) which can be 0 L x as much as several nanometers thick. Al2O3 This layer could cause a problem in making electrical contacts with outlets, for example, since it presents a barrier of roughly 10 eV to the flow of electrons in and out of the Al. Your requirement is that your transmission coefficient across any contact must be T > 10-10, or else the resistance will be too high for the high currents you’re using, causing a fire risk. Should you use aluminum wiring or not? (You can neglect G here.)
8. Nature 434, 361 - 364 (17 March 2005) Current measurement by real-time counting of single electrons JONAS BYLANDER, TIM DUTY & PER DELSING ~40 fA ~80 fA ~120 fA Electrons that successfully tunnel through the 50 junctions are detected using a fast single electron transistor (SET).
9. a-Radiation: Example 1 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Polonium-212 atom (Z = 84), which presents a square barrier of width L = 9.4 x 10-15 m (9.4 fermi) and height Uo = 26 MeV. What is the tunneling probability for an alpha particle with energy ~ 9 MeV each time the particle hits the barrier? What is the approximate lifetime of the 212Po?
10. Lecture 9, exercise 2 We just looked at Polonium, with an effective barrier width of ~10 fermi, and saw a tunneling probability of ~10-15. Now consider Uranium, which has a similar barrier height, but an effective width of about ~20 fermi. Estimate the tunneling probability in Uranium: a. 10-30 b. 10-14 c. 10-7
11. a-Radiation: Example 2 Consider a very simple model of a-radiation: Assume the alpha particle (m = 6.64 x 10-27 kg) is trapped in the nucleus of a Uranium-235 atom (Z = 92), which presents a square barrier of width L ~ 21 x 10-15 m (21 fermi) and height Uo = 28 MeV. What is the tunneling probability for an alpha particle with energy ~4.5 MeV each time the particle hits the barrier? What is the approximate lifetime of the 235U? [For this order of magnitude calculation you may neglect G.]
12. FYI: The sun! The solar nuclear fusion process starts when two protons fuse together (they eventually become a helium nucleus, which fuses with another one, releasing two energetic protons). The temperature of the sun is some 1015 K. This corresponds to an average kinetic energy: -16 3/2 kBT = 3 x 10 J (kB = Boltzman’s constant) How much energy would they need classically to “touch” (approach to a distance the size of the nucleus, ~1 fm)? They are both charged, and therefore repel each other: 2 9 -19 2 -15 -13 U(r) = (1/4 e0)xe /r = (9x10 )x(1.6x10 C) /10 m = 2 x 10 J Thus, classically, the protons in the sun do not have enough energy to overcome their coulomb repulsion. How do they fuse then? By tunneling through the coulomb barrier!
13. Application: “Tunneling” & the NH3 Maser Question 1: What are the two lowest eigenstates of this ‘double-well’ potential? First consider two separate U(x) wells: U(x) 0 x H.O. potential  (x,t)  (x) x Ground state x The double-well states are superpositions x of these two single-well wavefunctions.
14. Application: “Tunneling” & the NH3 Maser Question 2: Given the energy difference between the ground and first excited states, E2 - E1 = 1.8x10-4 eV, estimate how long it U(x) takes for the N atom to “tunnel” E from one side of the NH molecule 2 3 E1 to the other? 0 x Two-well simulation Qmdbw
15. Application: “Tunneling” & the NH3 Maser Question 3: (Stimulated) Emission of radiation between these two lowest energy configurations of ammonia, E1 =0.004637 eV and E2 =0.004655 eV (DE = 1.8x10-4 eV) was used to create U(x) the Ammonia Maser, by C. Townes in E2 1954 (for which he won the Nobel E1 prize in 1964). What wavelength of 0 x radiation does the maser emit? Microwave Laser! (Maser)
16. Application: “Tunneling” & the NH3 Maser Astronomical N Masers: Plane of hydrogen atoms. U(x) E2 E1 0 x
17. Lecture 9, exercise 3 You are trying to make a laser that emits violet light (400 nm), based on the transition an U0 electron makes between the ground and first- E1 excited state of a double quantum well as E shown. Your first sample emitted at 390 nm. 0 What could you modify to shift the wavelength to 400 nm? a. decrease the height of the barrier b. increase the height of the barrier c. decrease the width of the barrier The wavelength of the emitted photon was too low → the frequency of the photon was too high → the frequency of the electron oscillating between the left and right well was too high → the probability to “tunnel” was too high! You can reduce this by increasing the barrier height.
18. Lecture 9, exercise 3,cont. As we raise the height of the central barrier, the coupling between the two wells decreases. In the limit of an infinite barrier, it looks like two independent wells → same wavefunction curvature for both the symmetric (ground state) and anti- symmetric (1st excited state) wavefunctions → same kinetic energy, i.e., degenerate solutions. 0 1 1 0
19. Example Problem Suppose an electron of KE = 0.1 eV approaches a barrier. a) What is the wavelength of this electron? b) For what barrier height will it have a 50% chance of penetrating 1 nm into the forbidden region? What about 1 mm? Solution: a) In the allowed region: Use the De Broglie relation, p = h/l, and the kinetic E = h2/2ml2 = 1.505eVnm2/l2 energy E = p2/2m. = 0.1 eV (for an electron) l = 3.9 nm b) In the forbidden region: U is the unknown barrier height and E = 0.1 eV. The 2 in e-2KL results from probability being |Y|2. K = (2m(U-E)) / e-2KL = 1/2 Note that all that really matters is U-E. You can U = E + ( ln2/2L)2 / 2m also use h2/2m = 1.505eVnm2 for an electron. = 0.1 eV + 7.4 10-22 J L = 1 nm. = 0.105 eV L = 1 mm. Penetration by a significant distance isn’t or = 0.1 eV + 7.4 10-28 J = 0.100000005 eV possible unless the energy is nearly allowed.