Quantum Physics. When we consider the motion of objects on the atomic level, we find that our classical approach does not work very well.

Transcription

1 Quantum Physics When we consider the motion of objects on the atomic level, we find that our classical approach does not work very well. For example, quantum physics describes how electrons surround the nucleus of the atom and other subatomic actions. Therefore, for understanding motion on the microscopic scale we must use Quantum Mechanics. "I think I can safely say that nobody understands quantum mechanics." - Richard P. Feynman "I don't like it, and I'm sorry I ever had anything to do with it" - Erwin Schrödinger "Anyone who is not shocked by the quantum theory has not understood it." - Niels Bohr

2 We know some small particles like electrons, protons, and neutrons. Quantum physics even describes the particles which make these particles! (The model of an atom that you were taught in high-school is a approximation). The electrons don't orbit like planets; they form blurred clouds of probabilities around the nucleus. Protons and neutrons? They're each made of three quarks, each with its own 'flavor' and one of three 'colors'. Lets not forget the gluons, the even smaller particles that hold this mess together when they collect and form glueballs. The quantum model of the atom is much more complex than the traditional model. The world of subatomic particles is a very bizarre one, filled with quantum probabilities and organized chaos. For example, the exact position and velocity of an electron is very hard to find because attempts to "see" it involve bouncing other particles off of it. By doing this, you've just changed the electron's velocity, so your data is useless. What quantum physics does is give us the statistical probability of the electron's location at any one moment.

3 Blackbody Radiation One of the earliest indications that classical physics was incomplete came from attempts to describe blackbody radiation. A black body is a theoretical object that absorbs 100% of the radiation that hits it. Therefore it reflects no radiation and appears perfectly black. Blackbody radiation is the emission of electromagnetic waves from the surface of an object. The distribution of blackbody radiation depends only the temperature of the object and is independent of the material.

4 The Blackbody Distribution The intensity spectrum emitted from a blackbody has a characteristic shape. The maximum of the intensity is found to occur at a wavelength given by Wien s Displacement Law: f peak = ( s -1 K -1 )T T = temperature of blackbody (K) Notice the plot is versus frequency on these plots, but since it is related to wavelength via, f λ = c λmaxt = i m K

5 Example: The solar radiation curve approximates closely a blackbody at 5900K. What region of the spectrum is 3 the sun most intense? = 491.5nm Consequences: What part of the spectrum is this? 3 T=300K λmax = microns 300 K What portion of the spectrum is this? 3 T=3 million K λmax 6 =9.667Å = 0.97nm K What portion of the spectrum is this? 3 T=3K λmax =9.667mm 3K What portion of the spectrum is this? λ max mk i = = K mk i = = mk i = = mk i = = m m 3 m m

6 The Ultraviolet Catastrophe Classical physics can describe the shape of the blackbody spectrum only at long wavelengths. At short wavelengths there is complete disagreement. This disagreement between observations and the classical theory is known as the ultraviolet catastrophe.

7 Planck s Solution In 1900, the German physicist Max Planck was able to explain the observed blackbody spectrum by assuming that it originated from oscillators on the surface of the object and that the energies associated with the oscillators were discrete or quantized: E n = nhf n = 0, 1, 2, 3 n is an integer called the quantum number h is Planck s constant: J s = ev s f is the frequency Planck s idea of quanta was the beginning of quantum physics. The smallest packet of energy a electron can give or absorb. (Like you can t take an elevator to the 32 nd ½ floor). With electrons any energy is a whole number of quantas. The correct mathematical work to Planck s work was later done by Erwin Schroedinger, an Austrian scientist.

8 Quantum Mechanics The essence of quantum mechanics is that certain physical properties of a system (like the energy) are not allowed to be just any value, but instead must be only certain discrete values.

9 Quantization of Light Einstein proposed that light itself comes in chunks of energy, called photons. Light is a wave, but also a particle. The energy of one photon is E = hf where f is the frequency of the light and h is Planck s constant. Useful energy unit: 1 ev = J

10 Example (a) Find the energy of 1 (red) 650 nm photon. (b) Find the energy of 100 (red) 650 nm photons. E = hf = J s m

11 Example A 200 W infrared laser emits photons with a wavelength of m while a 200 W ultraviolet laser emits photons with a wavelength of m. (a) Which has greater energy, a single infrared photon or a single ultraviolet photon? E = hf = hc λ The UV photon has the greater energy; its wavelength is smallest. (b) What is the energy of a single infrared photon and the energy of a single ultraviolet photon? E E UV IR = = hc λ hc λ UV IR = = J J = 18 ev = 0.62 ev

12 (c) How many photons of each kind are emitted per second? number of photons emitted per second = energy emitted/sec energy/photon For both lasers the energy emitted per second is 200 J. The UV laser emits photons/sec and the IR laser emits photons per second.

13 The PhotoElectric Effect When light is incident on a surface (usually a metal), electrons can be ejected. This is known as the photoelectric effect. Around the turn of the century, observations of the photoelectric effect were in disagreement with the predictions of classical wave theory.

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15 Experiments show: 1. Brighter light causes more electrons to be ejected, but not with more kinetic energy. 2. The maximum KE of ejected electrons depends on the frequency of the incident light. 3. The frequency of the incident light must exceed a certain threshold, otherwise no electrons are ejected. 4. Electrons are ejected with no observed time delay regardless of the intensity of the incident light.

16 1. Brighter light causes more electrons to be ejected, but not with more kinetic energy. Wave theory predicts a more intense beam of light, having more energy, should cause more electrons to be emitted and they should have more kinetic energy. Particle theory predicts a more intense beam of light to have more photons so more electrons should be emitted, but since the energy of a photon does not change with beam intensity, the kinetic energy of the ejected electrons should not change.

17 2. The maximum KE of ejected electrons depends on the frequency of the incident light. Wave theory cannot explain the frequency dependence of the maximum kinetic energy. Particle theory predicts the maximum kinetic energy of the ejected electrons to show a dependence on the frequency of the incident light. Each electron in the metal absorbs a whole photon: some of the energy is used to eject the electron and the rest goes into the KE of the electron.

18 3. The frequency of the incident light must exceed a certain threshold, otherwise no electrons are ejected. Wave theory can offer no explanation. Particle theory predicts a threshold frequency is needed. Only the incident photons with f>f threshold will have enough energy to free the electron from the metal.

19 4. Electrons are ejected with no observed time delay regardless of the intensity of the incident light. Wave theory predicts that if the intensity of the light is low, then it will take some time before an electron absorbs enough energy to be ejected from the metal. Particle theory predicts a low intensity light beam will just have a low number of photons, but as long as f>f threshold an electron that absorbs a whole photon will be ejected; no time delay should be observed.

20 First Photoelectric Experiment: Photoelectrons stopped by stopping voltage V stop. The kinetic energy of the most energetic photoelectrons is K max = ev stop K max does not depend on the intensity of the light! single photon ejects each electron Second Photoelectric Experiment: Photoelectric effect does not occur if the frequency is below the cutoff frequency f o, no matter how bright the light! single photon with energy greater than work function Φ ejects each electron slope = ab bc

21 The Photoelectric Effect Photoelectric Equation The previous two experiments can be summarized by the following equation, which also expresses energy conservation hf = K max + Φ (photoelectric equation) Using h Kmax = evstop Vstop = f e Φ e equation for a straight with slope h/e and intercept Φ/e h ab 2.35 V 0.72 V = = = = e bc Hz -15 slope V s Multiplying this result by e ( ) ( )( ) h = = V s C J s 38-

22 The Photoelectric Effect Explained (Einstein 1905, Nobel Prize 1921) The photoelectric effect can be understood as follows: Electrons are emitted by absorbing a single photon. A certain amount of energy, called the work function, Φ, is required to remove the electron from the material. The maximum observed kinetic energy is the difference between the photon energy and the work function. K max = E - Φ E = photon energy "just shows emission" E = Φ means K max is zero Cutoff Frequency f 0 = Φ/h

23 The work function is a property of the individual metal Element Work Function (ev) Element Work Function (ev) Aluminum 4.3 Nickel 5.1 Carbon 5.0 Silicon 4.8 Copper 4.7 Silver 4.3 Gold 5.1 Sodium 2.7 Application: Photocells, Solar Cells (5eV = 248nm) If the energy of a photon is high enough it can break bonds in molecules/materials

24 Zinc and cadmium have photoelectric work functions given by Φ Zn = 4.33 ev and Φ Cd = 4.22 ev, respectively. (a) If both metals are illuminated by UV radiation of the same wavelength, which one gives off photoelectrons with the greater maximum kinetic energy? Explain. (b) Calculate the maximum kinetic energy of photoelectrons from each surface if λ = 275 nm. K max = E Φ = h c / λ 4.33 ev J/eV = J Planck s constant: J s = ev s

25 Example The photoelectric threshold frequency of silver is Hz. What is the minimum energy required to remove an electron from silver? KE max = φ = = = hf hf threshold threshold ( 34 )( Js Hz) φ = 0 19 J = 4.30 ev

26 Example Two different monochromatic light sources, one yellow (580 nm) and one violet (425 nm), are used in a photoelectric effect experiment. The metal surface has a photoelectric threshold frequency of Hz. (a) Are both sources able to eject photoelectrons from the metal? Explain. The frequency of each source is f f yellow violet = = c λ c λ yellow violet = = Hz Hz. Only the violet light is above the threshold frequency.

27 Example continued: (b) How much energy is required to eject an electron from the metal? KE max = φ = = = hf hf threshold threshold ( 34 )( Js Hz) φ = 0 19 J = 2.56 ev

28 X-ray Production When high energy electrons impact a target x-ray photons can be emitted as the electrons are slowed. This process is called Bremsstrahlung (German for breaking radiation). There is a continuous spectrum of radiation emitted up to a cutoff frequency. The spikes in the spectrum are called characteristic x-rays. These peaks depend on the target material.

29 The Mass and Momentum of a Photon Photons have momentum, but no mass. We cannot use the formula p = mv to find the momentum of the photon. Instead: hf p = = c h λ

30 Photons Have Momentum, Compton Shift p = hf c h = (photon momentum) λ

31 Photons Have Momentum, Compton shift hf = hf ' + K Conservation of energy Since electron may recoil at speed approaching c we must use the relativistic expression for K 2 K mc γ ( 1) = 1 γ = 1 vc ( ) 2 where γ is the Lorentz factor Substituting K in the energy conservation equation f = c λ 2 h h hf = hf ' + mc ( γ 1 ) = + mc( γ 1) p = γ mv λ λ' h h Conservation of momentum along x = cosφ + γmv cosθ λ λ ' h Conservation of momentum along y 0 = sinφ γmvsinθ λ '

32 Photons Have Momentum, Compton shift cont d Want to find wavelength shift Δ λ = λ' λ Conservation of energy and momentum provide 3 equations for 5 unknowns (λ, λ, v, φ, and θ ), which allows us to eliminate 2 unknowns v and θ. Δ λ = h mc ( φ) 1-cos (Compton shift) λ, λ, and φ, can be readily measured in the Compton experiment h mc is the Compton wavelength and depends on 1/m of the scattering particle Loose end: Compton effect can be due to scattering from electrons bound loosely to atoms (m=m e peak at θ 0) or electrons bound tightly to atoms (m m atom >>m e peak at θ 0)

33 Example A photon is incident on an electron at rest. The scattered photon has a wavelength of 2.81 pm and moves at an angle of 29.5 with respect to the direction of the incident photon. (a) What is the wavelength of the incident photon? The Compton shift is Δλ = = = λ c ( 1 cosθ ) ( 2.43 pm)( 1 cos 29.5 ) pm. The incident wavelength is λ 0 = λ Δλ 1 = 2.81 pm pm = 2.50 pm.

34 Example continued: (b) What is the final kinetic energy of the electron? 55 kev J ) ( = = = = = λ λ λ λ hc h h c p p c K The final kinetic energy of the electron is equal to the change in the photon s energy.

35 The Wave Nature of Particles We have seen that light is described sometimes as a wave and sometimes as a particle. In 1924, Louis debroglie proposed that particles also display this dual nature and can be described by waves too! The debroglie wavelength of a particle is related to its momentum: λ= h/p (Use p = γm 0 v if the velocity is large.) That is why you can make an atom laser!

36 Electrons and Matter Waves If electromagnetic waves (light) can behave like particles (photons), can particles behave like waves? λ = h p (de Broglie wavelength) where p is the momentum of the particle Electrons θ 38-

37 We can investigate wave properties of particles through interference/diffraction

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39 Light as a Probability Wave How can light act both as a wave and as a particle (photon)? Standard Version: photons sent through double slit. Photons detected (1 click at a time) more often where the classical intensity: I 2 Erms cμ = is maximum. 0 Fig The probability per unit time interval that a photon will be detected in any small volume centered on a given point is proportional to E 2 at that point. Light is not only an electromagnetic wave but also a probability wave for detecting photons

40 Light as a Probability Wave, cont'd Single Photon Version: photons sent through double slit one at a time. First experiment by Taylor in We cannot predict where the photon will arrive on the screen. 2. Unless we place detectors at the slits, which changes the experiment (and the results), we cannot say which slit(s) the photon went through. 3. We can predict the probability of the photon hitting different parts of the screen. This probability pattern is just the two slit interference pattern that we discussed in Ch. 35. The wave traveling from the source to the screen is a probability wave, which produces a pattern of "probability fringes" at the screen. 38-

41 Light as a Probability Wave, cont'd Conclusions from the previous versions/experiments: 1. Light generated at source as photons. 2. Light absorbed at detector as photons. 3. Light travels between source and detector as a probability wave. 38-

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43 Schrödinger s Equation For light E(x, y, z, t) characterizes its wavelike nature, for matter the wave function Ψ(x, y, z, t) characterizes its wavelike nature. Like any wave, Ψ(x, y, z, t) has an amplitude and a phase (it can be shifted in time and or position), which can be conveniently represented using a complex number a+ib where a and b are real numbers and i 2 = -1. On the situations that we will discuss, the space and time variables can be grouped separately: ( xyzt,,, ) ψ (,,, ) Ψ = i t xyzte ω where ω=2πf is the angular frequency of the matter wave.

44 Schrödinger s Equation, cont d What does the wave function mean? If the matter wave reaches a particle detector that is small, the probability that that a particle will be detected there in a specified period of time is proportional to Iψ 2, where Iψ is the absolute value (amplitude) of the wave function at the detector s location. The probability per unit time interval of detecting a particle in a small volume centered on a given point in a matter wave is proportional to the value Iψ 2 at that point. Since ψ is typically complex, we obtain Iψ 2 by multiplying ψ by its complex conjugate ψ*. To find ψ* we replace the imaginary number i in ψ with i wherever it occurs. 2 ( )( ) ψ = ψψ* = a+ ib a ib

45 Schrödinger s Equation, cont d How do we find (calculate) the wave function? Matter waves are described by Schrödinger s equation. Light waves a described Maxwell s equations, matter waves are described by Schrödinger s equation. For a particle traveling in the x direction through a region in which forces on the particle cause it to have a potential energy U(x), Schrödinger s equation reduces to: 2 2 d ψ 8π m E U ( x ) ψ =0 (Schrodinger's eq. in 1D) dx h where E is the total energy of the particle. If U(x) = 0, this equation describes a free particle. In that case the total energy of the particle is simple its kinetic energy (1/2)mv 2 and the equation becomes: + ψ=0 + 2 π ψ=0 dx h 2 dx h d ψ 8π m mv d ψ p

46 Schrödinger s Equation, cont d In the previous equation, since λ=h/p, we can replace the p/h with 1/λ, which in turn is related to the angular wave number k=2π/λ. 2 d ψ dx + 2 k ψ =0 (Schrodinger's eq., free particle) 2 The most general solution is: ψ ( x)= Ae + Be ikx ikx Leading to: ω (, ) ψ ( ) = ( ) Ψ x t = x e Ae + Be e i t ikx ikx iωt ( ω ) ( + ω ) ikx t ikx t = Ae + Be 38-

47 Finding the Probability Density Iψ 2 iθ iθ e = cosθ + isin θ and e = cosθ isin θ Choose arbitrary constant B=0 and let A= ψ 0 ψ ( x) = ψ0 e ikx ( ) 2 ikx 2 2 ikx = 2 0e = 0 e ψ ψ ψ ikx 2 ikx ikx ikx ikx ikx ikx ( )( ) e = e e * = e e = e = 1 ( ) 2 ikx e = 0 = 0 ψ = ψ ψ 1 ψ (a constant) Fig

48 Heisenberg Uncertainty Principle To describe a particle a physicist would refer four properties, the position of the electron, its momentum, its energy, and the time. Heisenberg showed that no matter how accurate the instruments used, quantum mechanics limits the precision when two properties (let s built two pairs: momentum-position or energy-time) are measured at the same time. "The more precisely the POSITION is determined, the less precisely the MOMENTUM is known" WERNER HEISENBERG ( )

49 Heisenberg s Uncertainty Principle In the previous example, the momentum (p or k) in the x-direction was exactly defined, but the particle s position along the x-direction was completely unknown. This is an example of an important principle formulated by Heisenberg: Measured values cannot be assigned to the position r and the momentum p of a particle simultaneously with unlimited precision. Δx Δp where = h 2π Δy Δp Δz Δp x y z (Heisenberg's uncertainty principle) Note that if Δ p =Δ p =Δ p = 0 x y z Then Δx, Δy, and Δz 38-

50 Determinism of Classical Mechanics Suppose the positions and speeds of all particles in the universe are measured to sufficient accuracy at a particular instant in time It is possible to predict the motions of every particle at any time in the future (or in the past for that matter) An intelligent being knowing, at a given instant of time, all forces acting in nature, as well as the momentary positions of all things of which the universe consists, would be able to comprehend the motions of the largest bodies of the world and those of the smallest atoms in one single formula, provided it were sufficiently powerful to subject all the data to analysis; to it, nothing would be uncertain, both future and past would be present before its eyes. Pierre Simon Laplace

51 Measuring the position of an electron Shine light on electron and detect reflected light using a microscope Minimum uncertainty in position is given by the wavelength of the light So to determine the position accurately, it is necessary to use light with a short wavelength

52 Measuring the momentum of an electron By Planck s law E = hc/λ, a photon with a short wavelength has a large energy Thus, it would impart a large kick to the electron But to determine its momentum accurately, electron must only be given a small kick This means using light of long wavelength!

53 Fundamental Trade Off Use light with short wavelength: accurate measurement of position but not momentum Use light with long wavelength: accurate measurement of momentum but not position h ΔxΔp = ( h _ bar) 2 π The more accurately you know the position (i.e., the smaller Δx is), the less accurately you know the momentum (i.e., the larger Δp is); and vice versa

54 Heisenberg s Uncertainty Principle involving energy and time ΔEΔt h 2π = The more accurately we know the energy of a body, the less accurately we know how long it possessed that energy The energy can be known with perfect precision (ΔE = 0), only if the measurement is made over an infinite period of time (Δt = )

55 Barrier Tunneling As puck slides up hill, kinetic energy K is converted to gravitational potential energy U. If the puck reaches the top its potential energy is U b. The puck can only pass over the top if its initial mechanical energy E> U b. Otherwise the puck eventually stops its climb up left side of hill and slides back top left. For example, if U b =20 J and E=10 J, the puck will not make pass over the hill, which acts as a potential barrier. Fig

56 Barrier Tunneling, cont d What about an electron approaching an electrostatic potential barrier? Fig Due to the nature of quantum mechanics, even if E< U b there is a non-zero transmission probability (transmission coefficient T) that the electron will get through (tunnel) to the other side of the electrostatic potential barrier! ( ) 2 2bL 8π mub E where 2 T e b= h Fig Fig

57 The Scanning Tunneling Microscope (STM) As tip is scanned laterally across the surface, the tip is moved up or down to keep the tunneling current (tip to surface distance L) constant. As a result the tip maps out the contours of the surface with resolution on the scale of 1 nm instead of >300 nm for optical microscopes! 38-