Quantum Physics PDF

Chapter 2 QUANTUM PHYSICS OBJECTIVES: To learn certain experimental results that can be understood only by particle theory of electromagnetic waves. To learn the particle properties of waves and the wave properties of the...

Chapter 2 QUANTUM PHYSICS OBJECTIVES: To learn certain experimental results that can be understood only by particle theory of electromagnetic waves. To learn the particle properties of waves and the wave properties of the particles. To understand the uncertainty principle. Department of Physics - MIT, Manipal 1  Origin of Blackbody Kirchhoff and Bunsen used spectrum analysis to study the composition of the sun and they discovered the elements cesium (1860) and rubidium (1861). With the use of a spectroscope they had invented together, they managed to identify these two alkali metals that the world had no previous knowledge of. Their discoveries marked the beginning of a new era, introducing a new way to look for undiscovered elements. Kirchhoff was the first to explain the dark lines in the sun’s spectrum stating that they were caused by cooler gases in the sun’s atmosphere absorbing particular wavelengths of sunlight. These dark-lined spectra are now called absorption spectra. Department of Physics - MIT, Manipal 2  For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect Department of Physics - MIT, Manipal 3 black-body emissive power”  What is Blackbody Spectra In 1895, at the University of Berlin, Wien and Lummer punched a small hole in the side of a closed oven, and began to measure the radiation coming out. The beam coming out of the hole was passed through a diffraction grating, which sent the different wavelengths/frequencies in different directions, all towards a screen. A detector was moved up and down along the screen to find how much radiant energy was being emitted in each frequency range. They found a radiation intensity/frequency curve close to this. Department of Physics - MIT, Manipal 4 Blackbody Radiation and Planck’s Hypothesis The electromagnetic radiation emitted by the black body is called black-body radiation. Basic laws of radiation (1) All objects emit radiant energy. (2) Hotter objects emit more energy (per unit area) than colder objects. (Stefan’s Law ) P =  A e T4 (3) The peak of the wavelength distribution shifts to shorter wavelengths as the black body temperature increases. (Wien’s Displacement Law) λm T = constant Department of Physics - MIT, Manipal 5 Department of Physics - MIT, Manipal 6 (4) Rayleigh-Jeans Law: The intensity or power per unit area I (,T)d, emitted in the wavelength interval  to +d from a blackbody is given by 2  c kB T I(  , T ) = 4 It agrees with experimental measurements only for long wavelengths It predicts an energy output that diverges towards infinity as wavelengths become smaller and is called ultraviolet catastrophe. Department of Physics - MIT, Manipal 7 (5) Planck‘s Law: The intensity or power per unit area I (,T)d, emitted in the wavelength interval  to +d from a blackbody is given by 2  h c2 1 I(, T ) = 5 hc e λkT −1 Assumptions of this law are: Energy of an oscillator in cavity walls: En = n h f Amount of emission / absorption of energy will be integral multiples of hf. Department of Physics - MIT, Manipal 8 The results of Planck's law: The denominator [exp(hc/λkT)] tends to infinity faster than the numerator (λ–5), thus resolving the ultraviolet catastrophe and hence arriving at experimental observation: I (λ, T) → 0 as λ → 0. For very large λ, I (λ, T) → 0 as λ → . hc exp( hckT ) − 1   kT  I(  ,T ) → 2  c −4 k T From a fit between Planck's law and experimental data, Planck’s constant was derived to be h = 6.626 × 10–34 J-s. Department of Physics - MIT, Manipal 9 Photoelectric Effect Ejection of electrons from the surface of certain metals when it is irradiated by an electromagnetic radiation of suitable frequency is known as photoelectric effect. A E V C Photoelectric Effect (T – Evacuated glass/ quartz tube, E – Emitter Plate / Photosensitive material / Cathode, C – Collector Plate / Anode, V – Voltmeter, A - Ammeter) Department of Physics - MIT, Manipal 10 Classical Predictions Experimental Observations 1. Electrons ejection should be 1. No photoemission for frequency frequency independent. below threshold frequency 2. KE of the electrons should increase 2. KMAX is independent of light with intensity of light. intensity. 3. Measurable/ larger time interval 3. Instantaneous effect between incidence of light and 4. KE of the most energetic ejection of photoelectrons. photoelectrons is, KMAX = e Vs & 4. KMAX should not depend upon the it increases with increasing f. frequency of the incident light. Experimental results contradict classical predictions Department of Physics - MIT, Manipal 11 Einstein’s Interpretation of electromagnetic radiation: 1. Electromagnetic waves carry discrete energy packets (light quanta called photons now). 2. The energy E, per packet depends on frequency f: E = hf. 3. More intense light corresponds to more photons, not higher energy photons. 4. Each photon of energy E moves in vacuum at the speed of light: c = 3 x 108 m/s and each photon carries a momentum, p = E/c. Einstein’s photoelectric equation Kmax = hf −  Department of Physics - MIT, Manipal 12 Compton Effect When X-rays are scattered by free/nearly free electrons, they suffer a change in their wavelength which depends on the scattering angle. Classical Predictions: Effect of oscillating electromagnetic waves on electrons: (a) oscillations in electrons, re-radiation in all directions (b) radiation pressure - electrons accelerate in the direction of propagation of the waves Different electrons will move at different speeds after the interaction. The scattered wave frequency should show a distribution of Doppler-shifted values Department of Physics - MIT, Manipal 13 Schematic diagram of Compton’s apparatus Graph of scattered x-ray intensity versus wavelength Department of Physics - MIT, Manipal 14 Derivation of the Compton shift equation Photon is treated as a particle having energy E = hfo = hc/o and zero rest energy. Photons collide elastically with free electrons initially at rest as shown in figure. In the scattering process, the total energy and total linear momentum of the system must be conserved. o = wavelength of the incident photon po = h/o = momentum of the incident photon Eo = hc/o = energy of the incident photon ’ = wavelength of the scattered photon p’ = h/’ = momentum of the scattered photon E’ = hc/’ = energy of the scattered photon Department of Physics - MIT, Manipal 15 Conservation of energy: Eo = E’ + K Conservation of momentum: x-component: po = p′ cos θ + p cos ϕ y-component: 0 = p′ sin θ − p sin ϕ Relativistic equations: v = speed of the electron m = mass of the electron 1 p =  m v = momentum of the electron where  = v2 1− c2 E= p2 c 2 + m2 c 4 = total relativistic energy of the electron K = E − m c2 = kinetic energy of the electron By using above relations and simplifying, we will get, 𝐡 Compton shift 𝛌′ − 𝛌𝐨 = 𝟏 − 𝐜𝐨𝐬 𝛉 𝐦𝐜 Department of Physics - MIT, Manipal 16 Photons and Electromagnetic Waves [Dual Nature of Light] Light exhibits diffraction and interference phenomena that are only explicable in terms of wave properties. Photoelectric effect and Compton Effect can only be explained taking light as photons / particle. This means true nature of light is not describable in terms of any single picture, instead both wave and particle nature have to be considered. In short, the particle model and the wave model of light complement each other. Department of Physics - MIT, Manipal 17 1. A sodium surface is illuminated with light having a wavelength of 300 nm. The work function for sodium metal is 2.46 eV. Find A. The maximum kinetic energy of the ejected photoelectrons and B. The cutoff wavelength for sodium 2. Molybdenum has a work function of 4.2eV. (a) Find the cut off wavelength and cut off frequency for the photoelectric effect. (b) What is the stopping potential if the incident light has wavelength of 180 nm? 3. X-rays of wavelength o = 0.20 nm are scattered from a block of material. The scattered X-rays are observed at an angle of 45° to the incident beam. Calculate their wavelength. What if we move the detector so that scattered X-rays are detected at an angle larger than 45°? Does the wavelength of the scattered X-rays increase or decrease as the angle increase? Department of Physics - MIT, Manipal 18 4. A-rays of wavelength o = 0.20 nm are scattered from a block of material. The scattered X-rays are observed at an angle of 45° to the incident beam. Calculate their wavelength. What if we move the detector so that scattered X-rays are detected at an angle larger than 45°? Does the wavelength of the scattered X-rays increase or decrease as the angle increase? Department of Physics - MIT, Manipal 19 de Broglie Hypothesis - Wave Properties of Particles Wavelength associated with particle of mass m moving with velocity v is given by 𝒉 𝒉 de Broglie wavelength: 𝝀 = = 𝒑 𝒎𝒗 The momentum (p) of an electron accelerated through a potential difference of V is 𝑝 = 𝑚𝑣 = 2 𝑚 𝑒 ∆𝑉 𝐸 Frequency of the matter wave associated with the particle is , where E is total ℎ relativistic energy of the particle. 𝐸 f= ℎ Department of Physics - MIT, Manipal 20 Davisson and Germer Experiment (Experimental verification of de-Broglie hypothesis) Assumption: electron as wave The idea is to determine the wavelength of electron using the Bragg’s diffraction law and then compare it with the de-Broglie’s wavelength. A beam of electron is produced by a heated filament and accelerated by potential 𝑉. ( Here 𝑉 = 54 V) This beam of electron is then scattered by a nickel crystal. Intensities of the scattered electrons are measured as a function angle 𝜙 (𝜙 is the angle between incident beam and scattered beam) Department of Physics - MIT, Manipal 21  The Bragg’s diffraction law states that 𝑑 sin 𝜙 = 𝑛λ Here 𝑑 is the inter-atomic spacing in nickel, and is equal to 0.215 𝑛𝑚. 𝑛 = 1, for the first diffraction maximum, which is at 𝜙 = 50° Thus on substituting these experimental numbers we obtain for the electron wavelength as λ = 0.165 nm Now let’s calculate the electron wavelength according to de-Broglie’s hypothesis. From conservation of energy we have 1 𝑚𝑣 2 = 𝑒 𝑉, where 𝑣 is the velocity of electron. 2 Department of Physics - MIT, Manipal 22  Momentum of the electron is given by 𝑝 =𝑚𝑣 = 2𝑚𝑒𝑉 The wavelength therefore is ℎ ℎ ℎ 𝜆= = = 𝑝 𝑚𝑣 2𝑚𝑒𝑉 On substitution, for 𝑉 = 54 V, we obtain λ = 0.167 nm We see that wavelength derived above agrees well with that from the Bragg’s diffraction law. This proves the de-Broglie’s hypothesis. Department of Physics - MIT, Manipal 23 The Quantum Particle If we add up large number of waves such that constructive interference takes place in small localized region of space a wavepacket, which represents a quantum particle can be formed. Superposition of two waves Wave packet Department of Physics - MIT, Manipal 24 Mathematical representation of a wave packet: 𝑦1 = 𝐴 𝑐𝑜𝑠(𝑘1 𝑥 − 𝜔1 𝑡) and 𝑦2 = 𝐴 𝑐𝑜𝑠(𝑘2 𝑥 − 𝜔2 𝑡) where 𝑘 = 2𝜋/𝜆 , 𝜔 = 2𝜋𝑓 The resultant wave y = y1 + y2 𝛥𝑘 𝑦 = 2𝐴 𝑐𝑜𝑠 2 𝑥 − 𝛥𝜔 2 𝑡 𝑐𝑜𝑠 𝑘1 +𝑘2 2 𝑥 − 𝜔1+𝜔 2 2𝑡 where k = k1 – k2 and  = 1 – 2. Phase speed, the speed with which Group speed, the speed of the wave wave crest of individual wave moves, is packet, is given by given by 𝛥𝜔 2 𝛥𝜔 𝜔 𝑣𝑔 = 𝛥𝑘 = 𝑣𝑝 = 𝑓 𝜆 or 𝑣𝑝 = 𝛥𝑘 𝑘 2 Department of Physics - MIT, Manipal 25 Relation between group speed (vg) and phase speed (vp): 𝜔 𝑣𝑃 = = 𝑓𝜆  𝜔 = 𝑘 𝑣𝑃 𝑘 𝑑𝜔 𝑑(𝑘𝑣𝑃 ) 𝑑𝑣𝑃 But 𝑣𝑔 = = = 𝑘 + 𝑣𝑃 𝑑𝑘 𝑑𝑘 𝑑𝑘 Substituting for k in terms of λ, we get 𝒅𝒗𝑷 𝒗𝒈 = 𝒗𝑷 − 𝝀 𝒅𝝀 Relation between group speed (vg) and particle speed (u): 𝐸 2𝜋 2𝜋 2𝜋𝑝 𝜔 = 2𝜋𝑓 = 2𝜋 and 𝑘 = = = ℎ 𝜆 ℎ Τ𝑝 ℎ 2𝜋 𝑑𝜔 ℎ 𝑑𝐸 𝑑𝐸 𝑣𝑔 = = 2𝜋 = 𝑑𝑘 𝑑𝑝 𝑑𝑝 ℎ For a classical particle moving with speed u, the kinetic energy E is given by 𝑝2 2 𝑝 𝑑𝑝 𝑑𝐸 𝑝 𝐸 = 1 2 𝑚 𝑢2 = and 𝑑𝐸 = or = = 𝑢 2𝑚 2𝑚 𝑑𝑝 𝑚 𝒅𝝎 𝒅𝑬 𝒗𝒈 = = = 𝒖 𝒅𝒌 𝒅𝒑 Department of Physics - MIT, Manipal 26 Double–Slit Experiment Revisited (a) Schematic of electron beam interference experiment, (b) Photograph of a double-slit interference pattern produced by electrons 𝒅 𝒔𝒊𝒏 𝜽 = 𝒎 𝝀 , where m is the order number and λ is the electron wavelength. The electrons are detected as particles at a localized spot on the detector screen at some instant of time, but the probability of arrival at the spot is determined by finding the intensity of two interfering waves. Department of Physics - MIT, Manipal 27 Uncertainty Principle Heisenberg uncertainty principle: It is fundamentally impossible to make simultaneous measurements of a particle’s position and momentum with infinite accuracy. ( x ) ( px) ≥ h / 4 One more relation expressing uncertainty principle is related to energy and time which is given by ( E ) ( t ) ≥ h / 4 Department of Physics - MIT, Manipal 28 The radius of our Sun is 6.96 x 108 m, and its total power output is 3.77 x 1026 W. (a) Assuming that the Sun’s surface emits as a black body, calculate its surface temperature. (b) Using the result, find max for the Sun. Ans: 5750 K, 504 nm Wien’s displacement law -> 2.898 x 10-3 m – K Stefan constant  = 5.67 x 10-8 Wm Department of Physics - MIT, Manipal 29 A blackbody at 7500 K consists of an opening of diameter 0.050 mm, looking into an oven. Find the number of photons per second escaping the hole and having wavelengths between 500 nm and 501 nm. Ans: 1.30 x 1015/s Department of Physics - MIT, Manipal 30 The stopping potential for photoelectrons released from a metal is 1.48 V larger compared to that in another metal. If the threshold frequency for the first metal is 40.0 % smaller than for the second metal, determine the work function for each metal. Ans: 3.70 eV, 2.22 eV Department of Physics - MIT, Manipal 31 A 0. 00160 nm photon scatters from a free electron. For what photon scattering angle does the recoiling electron have kinetic energy equal to the energy of the scattered photon? Ans: 70° Department of Physics - MIT, Manipal 32 (a) An electron has a kinetic energy of 3.0 eV. Find its wavelength. (b) Also find the wavelength of a photon having the same energy. Ans: 7.09 x 10–10 m, 4.14 x 10–7 m Department of Physics - MIT, Manipal 33 The lifetime of an excited atom is given as 1.0 x 10-8 s. Using the uncertainty principle, compute the line width f produced by this finite lifetime? Ans: 8.0 x 106 Hz Department of Physics - MIT, Manipal 34 Use the uncertainty principle to show that if an electron were confined inside an atomic nucleus of diameter 2 x 10–15 m, it would have to be moving relativistically, while a proton confined to the same nucleus can be moving nonrelativistically. Ans: vELECTRON  0.99996 c, vPROTON  1.8 x 107 m/s Department of Physics - MIT, Manipal 35