Introduction to Quantum Mechanics: Planck's Constant, Energy Levels, and Tunneling - Prof., Study notes of Electrical and Electronics Engineering

Download Introduction to Quantum Mechanics: Planck's Constant, Energy Levels, and Tunneling - Prof. and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! 1 Chapter 2 Introduction to Quantum Mechanics In order to understand the current-voltage characteristics of a semiconductor, we need some knowledge of the electron behavior in a crystal when subjected to various potential functions. Section 2.1 Principles of Quantum Mechanics Energy Quanta: Experiments with the photoelectric effect and dark body radiation show that the energy comes in discrete packets. E = hν where ν = frequency h = Planck’s constant = 6.6262 x 10-34 J sec E is in units of Joules and can be expressed in electron volts (eV). Thus we frequently interchange volts with electron volts to mean the same energy. See appendix F on page 721. If an electron is accelerated through a potential of 1 volt, then the energy is T (kinetic energy) = 1 e V = 1.602 x 10-19 J. 1 eV = 1.602 x 10-19 J The text uses h = 6.625 x 10-34 J sec instead of the more exact value. Work Function The minimum energy to remove an electron from a material is called the work function. This minimum energy is supplied by a photon of frequency ν0. Thus the maximum kinetic energy (T in this text) of the freed electron is: Tmax = ½ (mass) velocity2 = ½ mv2 = hν – hν0 where ν ≥ ν0 This is the energy to liberate an electron from a bound state in an atom or molecule and move it to free space. The electron does not have excess energy in the form of momentum at this min value. Exercise 2.1 page 29 Determine the energy of a photon at a particular wavelength. a) λ = 10,000 Å where 1 Å = 10-10 m = 0.1 x 10-9 m = 10-8 cm This is in the far infrared wavelengths. E hν= 19 19 19 1.988 101.988 10 1.602 10 JIn terms of eV J C − − − × → × = × E hν= c λν= cfreq ν λ = hc λ = 34 8 10 (6.625 10 sec)(3 10 ) 10,000 10 mJ s m − − × ⋅ × = × 191.988 10 J−= × 1.241 eV= Exercise 2.1 b) λ = 10 Å = 1 nm “a thousand times shorter λ” This is in the x-ray wavelengths. 34 8 10 (6.625 10 sec)(3 10 ) 10 10 mJ s m − − × ⋅ × = × 16 16 19 1.988 101.988 10 1.602 10 JIn terms of eV J C − − − × → × = × E hν= c λν= cfreq ν λ = E hν= hc λ = 161.988 10 J−= × 1,241 eV= Light Wave-Particle Duality Waves exhibit particle-like behavior (photoelectric effect). Light waves at only a certain minimum wavelength will liberate electrons in the above device. 2 Light Wave-Particle Duality Particles exhibit wave-like behavior (diffraction, interference). A stream of particles (photons) in a light beam can create an interference pattern that depends on the energy per photon. Photon Momentum The momentum of a photon is stated in terms of energy since the photon is “massless”. The momentum is given by: λ hp = de Broglie Wavelength In terms of the duality of light, de Broglie hypothesized that the wavelength of a particle can be expressed by: de Broglie Wavelength Note that these relationships are for VERY small particles smaller than neutrons and protons. p h =λ Exercise 2.2 page 29 a) Find the momentum and energy of a particle with mass = 5 x 10-31 kg de Broglie λ = 180 Å The energy of the particle is all kinetic in this case so: 2 21 2 2 p pE m m m ⎛ ⎞= =⎜ ⎟ ⎝ ⎠ 2 2 1energy kinetic the velmE ⋅== hp λ = pwhere p m vel vel m = ⋅ ⇒ = 34 10 6.625 10 180 10 J s m − − × ⋅ = × 263.6806 10 kg m s −= × 2 26 31 3.68 10 2(5 10 ) kg m s kg − − ⋅⎛ ⎞×⎜ ⎟ ⎝ ⎠= × 211.35 10 J−= × 38.46 10 eV−= × 263.6806 10 N s−= × ⋅ The Uncertainty Principle Heisenberg, 1927, showed that we cannot describe with absolute accuracy the behavior of subatomic particles. Thus we will describe the probabilities of particle behavior. It is impossible to simultaneously know precisely the position and momentum of a particle. Δp is the uncertainty in the momentum Δx is the uncertainty in the position Then Δp Δx ≥ ħ where ħ = h/2π = 1.054 x 10-34 J·sec The Uncertainty Principle It also is impossible to simultaneously know the exact energy of a particle at some specific instant of time t. ΔE is the uncertainty in the energy Δt is the uncertainty in the time Then ΔE Δt ≥ ħ where ħ = h/2π The physical meaning of this is shown by the exact position of an electron. We cannot know the exact position of the electron at time t. What we can develop is a probability density function of its position. 5 The Infinite Potential Well – Standing Wave Finally, the time-independent wave solution is given by: 3... 2, 1,n wheresin2)( =⎟ ⎠ ⎞ ⎜ ⎝ ⎛=Ψ a xn a x π A bound electron in the infinite potential wave is a standing wave as shown by the next slide (figure 2.6 on page 38). The energy of a bound particle in an infinite potential well is quantized. 2 222 2 energy where ma nEE n π == Figure 2.6 (a) four lowest discrete energy levels (b) Corresponding wave functions. (c) Corresponding probability functions. Exercise 2.5 on page 38 The width of the potential well a = 10 Å Calculate the first three energy levels in term of eV. 2 2 2 22n nE ma π = eVJJE 3743.010022.6)1)(10022.6( 202201 =×=×= −− eVJJE 497.110409.2)2)(10022.6( 192202 =×=×= −− eVJJE 368.3104198.5)3)(10022.6( 192203 =×=×= −− Electron in an infinite potential well. ( )234 2 2 31 10 2 1.054 10 2(9.11 10 )(10 10 ) J s n kg m π− − − × ⋅ = × × 20 2(6.022 10 )J n−= × Units J N m= ⋅ 2 kg m m s ⋅ = 2 2 2 J sso kg m ⇒ ⋅ 2J J = J=2 2 2 kg m s J ⋅ = The Step Potential Function A flux of particles is incident on the potential barrier traveling in the +x direction. Read the text for the derivation (pages 39-41) of the results that follow. Let the energy of the particles be less than this step potential barrier height. E < V0 The analysis shows: 1. An incident traveling wave of velocity vi 2. A reflected traveling wave of velocity vr 3. vi = vr velocity of incident and reflected waves are equal. 4. The reflection coefficient R = 1 The Step Potential Function Since the reflection coefficient R = 1 then all the particles are eventually reflected back into region I. So a particle may penetrate into region II at least for a while until it returns to region I as a reflection. Particles are neither absorbed nor transmitted into region II. Further analysis shows that the probability density function is not equal to zero in region II! Exercise 2.7 on page 42 The probability of finding a particle at a distance d in region II compared to that at x = 0 is given by dKe 22− Consider an electron on region 1 (traveling right) at a velocity v = 105 m/s. The potential barrier height = 3 E (kinetic energy) of the electron. Find the probability of finding the electron at a distance d = 10 Å into region II. 21 2 E m vel= ⋅ ( )231 51 (9.11 10 ) 10 /2 kg m s −= × 214.56 10 J−= × 6 Exercise 2.7 2 02 2 ( )mGiven K V E= − 31 21 34 2 2(9.11 10 ) (2)(4.56 10 ) (1.054 10 ) kg J J s − − − × = × × 22Probability K de−= 2 2 (3 )m E E= − Set the potential barrier height V0 to 3E. 2 2 (2 )m E= 9 -11.22 10 m = × 9 1 102(1.22 10 )(10 10 )m me − −− × ×= 0.0872= 8.72%= 2 kgUnits Js 22 2kg ms kg s⋅ = 2 1 m = 2 1 1so m m ⇒ = The Potential Barrier A flux of particles is incident on the potential barrier traveling in the +x direction. Let the energy of the particles be less than the barrier height. Eparticle < V0 The 3 region’s solutions are: 1Region for )( 11 111 xjKxjK eBeAx −+=Ψ 2Region for )( 22 222 xKxK eBeAx −+=Ψ 3Region for )( 11 333 xjKxjK eBeAx −+=Ψ The Potential Barrier 1Region for )( 11 111 xjKxjK eBeAx −+=Ψ 2Region for )( 22 222 xKxK eBeAx −+=Ψ 3Region for )( 11 333 xjKxjK eBeAx −+=Ψ 1 2 02 2 2 2 and ( )mE mWhere K K V E= = − Since there is nothing to cause a reflection in region III, the coefficient B3 = 0. The ratio of incident flux from region I that reaches region III is given by: * 11 * 33 AA AAT ⋅ ⋅ = The Potential Barrier The wave functions thru the potential barrier. When E << V0, the solution shows that there is a finite probability of a particle in region III. This phenomenon is called tunneling. aKe V E V ET 22 00 116 −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≈ where a = barrier width and T = the tunneling probability Roller Coaster Versus Quantum Tunneling Figures from Kasap, Principles of Electronic Material and Devices Roller coaster released from A cannot reach E in the Newtonian world. However, there is a chance of tunneling to the other side in quantum theory. Exercise 2.8 pg 45 Estimate the tunneling probability of an electron tunneling thru a rectangular barrier with: Barrier height of V0 = 1 eV Barrier width of a = 15 Å. The electron energy = 0.20 eV 9 102(4.595 10 / )(15 10 )0.20 0.2016 1 1 1 m me −− × ×⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ We need to find K2. 2 02 2 ( )mK V E= − 31 19 34 2 2(9.11 10 ) (1 0.20)(1.609 10 ) (1.054 10 ) kg− − − × = − × × 9 14.595 10 m−= × aKe V E V ET 22 00 116 −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ≈ 62.64 10−= × .000264%= 7 Section 2.4 Section 2.4 Extensions of the Wave theory to atoms will not be covered and is not testable. End of Chapter 2 Introduction to Quantum Mechanics