Quantum Physics PDF

Unit -II 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 th...

Unit -II 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, Bangalore 1 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS A body which absorbs all the electromagnetic radiation falling on it is called a black body. Black bodies are good absorbers of radiation and are also good emitters. A blackbody is one whose absorptivity is 100%. A black body will emit radiation at the fastest rate. A black body has maximum emissive power at a particular temperature. Blackbodies absorb and re-emit radiation in a characteristic pattern called a spectrum. Lamp black surface may be considered as perfectly black for all practical purposes. Department of Physics - MIT, Bangalore 2 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS DEFINITION OF A BLACK BODY An object that absorbs all incident radiation. A small hole cut into a cavity is the most popular and realistic example. None of the incident radiation escapes. It consists of a double walled hollow metallic sphere with a narrow opening and lamp-blacked. When the radiation enters into the body through the hole, it suffers multiple reflections inside the sphere and is completely absorbed. This causes a heating ofDepartment the cavity walls. of Physics - MIT, Bangalore 3 BTech-PHYSICS-INTRODUCTION TO QUANTUM PHYSICS 3 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS The oscillators in the cavity walls vibrate and cavity walls re-radiate at wavelengths corresponding to the temperature of the cavity, producing standing waves in the cavity. Some of the energy from these standing waves can leave through the opening. The electromagnetic radiation emitted by the black body is called black-body radiation. Standing waves is a wave that oscillates in time but whose peak profile does move in space. It is a combination of two waves moving in opposite directions, each having the same amplitude and frequency. The phenomenon is the result of interference; that is, when waves are superimposed, their energies are either added together or canceled out. In the case of waves moving in the same direction, interference produces a traveling wave. For oppositely moving waves, interference produces an oscillating wave fixed in space. Department of Physics - MIT, Bangalore 4 4 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS General characteristics of curves 1. At the given temperature the intensity of radiation increases along with the wavelength and reaches the maximum value at a particular value of wavelength λmax. Then it decreases along with the increase of the wavelength. 2. At a given temperature, the intensity is not uniformly Figure shows Intensity of blackbody distributed in the radiation radiation versus wavelength at three spectrum of a black body. temperatures. Department of Physics - MIT, Bangalore 5 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS General characteristics of curves 3. The wavelength λmax at the maximum emission of intensity shifts towards the shorter wavelength as the temperature increases. λmaxT= 2.898*10-3 m.K This represents Wien’s displacement law. Department of Physics - MIT, Bangalore 6 BLACKBODY RADIATION & PLANCK’S HYPOTHESIS General characteristics of curves 4. For all wavelengths the energy emission increases along with the temperature. 5. The area under each curve gives the amount of energy emitted at a given temperature. Department of Physics - MIT, Bangalore 7 Blackbody Radiation and Planck’s Hypothesis The electromagnetic radiation emitted by black body is called black-body radiation. Basic laws of radiation (1)All objects at any temperature emits electromagnetic waves in the form of thermal radiation. (2) The total power of the emitted radiation increases with temperature. (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 λm T = 2.898 × 10-3 m.K Department of Physics - MIT, Bangalore 8 (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 known as the ultraviolet catastrophe. Department of Physics - MIT, Bangalore 9 Which star is hotter? Department of Physics - MIT, Bangalore 10 (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 can have only discrete values: En = n h f Amount of emission / absorption of energy will be integral multiples of hf. Department of Physics - MIT, Bangalore 11  In the Rayleigh–Jeans model, the average energy associated with a particular wavelength of standing waves in the cavity is the same for all wavelengths and is equal to kBT. Planck used the same classical ideas as in the Rayleigh–Jeans model to arrive at the energy density as a product of constants and the average energy for a given wavelength. The average energy is not given by the equipartition theorem. A wave’s average energy is the average energy difference between levels of the oscillator, weighted according to the probability of the wave being emitted. This weighting is based on the occupation of higher-energy states as described by the Boltzmann distribution law Department of Physics - MIT, Bangalore 12 Planck’s model Department of Physics - MIT, Bangalore 13 The results of Planck's law: The denominator [exp(hc/λkT)] tends to zero 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, Bangalore 14 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 T 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, Bangalore 15 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, Bangalore 16 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, Bangalore 17 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 Doppler effect is the change in frequency of a wave for an observer moving relative to the source of the wave. Department of Physics - MIT, Bangalore 18 Schematic diagram of Compton’s apparatus Graph of scattered x-ray intensity versus wavelength Department of Physics - MIT, Bangalore 19 Diffraction of X-Rays by Crystals Crystal acts as 3D grating for X-rays. Condition for constructive interference (maxima in the reflected beam) is 2d sin  = m m = 1 , 2 , 3 ,... This condition is known as Bragg’s law 20 Department of Physics, MIT Bengaluru ▪ The wavelength was measured with a rotating crystal spectrometer using graphite (carbon) as the target. ▪ X- ray photons are scattered through 90° from a carbon target. ▪ The incident X-rays undergo scattering in many directions spanning through 0 to 1800. ▪ These scattered radiations were analyzed with a rotating crystal spectrometer. ▪ Intensity was measured with an ionization chamber that generated a current proportional to the intensity. ▪ At a given angle, only one frequency for scattered radiation is seen. ▪ The difference (λ’-λ0), (λ0 is wavelength of the incident radiation and λ’- wavelength of the scattered radiation) which indicates the enhancement in the wavelength, is called Compton shift. Compton shift (Δλ) is independent of the wavelength of the incident radiation. ▪ λ’ is independent of the target material. ▪ Δλ depends only on the scattering angle θ as the scattering angle is increased Δλ also increases. Department of Physics - MIT, Bangalore 21 Derivation of the Compton shift equation Photon is treated as a particle having energy E = hfo = hc/o. 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, Bangalore 22 Conservation of energy: Eo + mc2= E’ + Ee 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, Bangalore 23  From momentum conservation- po − p′ cos θ = p cos ϕ p′ sin θ = p sin ϕ SQUARING BOTH THE SIDES AND ADDING, po2 − 2pop′ cos θ + p′2 = p2 SUBSTITUTING THIS p2 IN THE EQUATION GOT BY CONSERVATION OF ENERGY: Eo − E′ 2 + 2 Eo − E′ mc 2 = p2 c 2 ONE GETS 2 + 2 E − E′ mc 2 = p 2 − 2p p′ cos θ + p′2 c 2 Eo − E′ o o o SUBSTITUTING PHOTON ENERGIES AND PHOTON MOMENTA ONE GETS 2 2 2 hc hc hc hc hc hc hc hc λo − λ′ +2 λo − λ′ mc 2 = λo −2 λo λ′ cos θ + λ′ Department of Physics - MIT, Bangalore 24 SIMPLIFYING ONE GETS 2 hc hc hc hc 2 1 1 λo − 2 λo 𝛌′ + 𝛌′ + 2 hc λo − 𝛌′ mc 2 2 hc hc hc hc 2 = λ −2 λ 𝛌′ cos θ + 𝛌′ o o hc hc ie, − + 1 − 1 λ′ mc 2 =− cos θ λo λ′ λo λo λ′ λ′ −λo hc OR, λo λ′ mc 2 = 1 − cos θ λo λ′ ′ h COMPTON SHIFT: λ − λo = 1 − cos θ mc Department of Physics - MIT, Bangalore 25 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, Bangalore 26  de Broglie Hypothesis - Wave Properties of Particles Louis de Broglie postulated that because photons have both wave and particle characteristics, perhaps all forms of matter have wave-like properties, with the wavelength λ related to momentum p in the same way as for light.Wavelength associated with particle of mass m moving with velocity v is given by de Broglie 𝒉 𝒉 wavelength: 𝝀 = = 𝒑 𝒎𝒗 The momentum (p) of an electron accelerated through a potential difference of V is 𝑝 = 𝑚𝑣 = 2 𝑚 𝑒 ∆𝑉 and its K.E. is ½ mv2 = eΔV 𝐸 Frequency of the matter wave associated with the particle is , where E is total ℎ 𝐸 relativistic energy of the particle. f=. ℎ Davisson-Germer experiment and G P Thomson’s electron diffraction experiment confirmed de Broglie relationship. Atomic beams, and beams of neutrons, also exhibit diffraction when reflected from regular crystals. Thus, de Broglie's formula Department of Physics - MIT, Bangalore seems to apply to any kind of matter. 27 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 28  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 29  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 30 The Quantum Particle To represent a quantum wave, we have to combine the essential features of both an ideal particle and an ideal wave. An essential feature of a particle is that it is localized in space. But an ideal wave is infinitely long (non-localized).To build a localized entity from an infinitely long wave, waves of same amplitude, but slightly different frequencies are superposed. If we add up large number of waves such that constructive interference takes place in small localized region of space is called wave packet, which represents a quantum particle can be formed. Superposition of two waves Wave packet Department of Physics - MIT, Bangalore 31 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, Bangalore 32 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 𝑑𝐸 𝑑𝐸 ℎ 𝑣𝑔 = = 2𝜋 = 𝑑𝑘 𝑑𝑝 𝑑𝑝 ℎ For a classical particle moving with speed u, the kinetic energy E is given by 1 2 𝑝2 2 𝑝 𝑑𝑝 𝑑𝐸 𝑝 𝐸 = 2 𝑚𝑢 = and 𝑑𝐸 = or = = 𝑢 2𝑚 2𝑚 𝑑𝑝 𝑚 𝒅𝝎 𝒅𝑬 𝒗𝒈 = = = 𝒖 𝒅𝒌 𝒅𝒑 To represent a realistic wave packet, confined to a finite region in space, we need the Department of Physics - MIT, Bangalore superposition of large number of harmonic waves with a range of k-values 33 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, Bangalore 34 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, Bangalore 35  Timeline: 1924 Erwin Early 1900 Planck 1905: Einstein said 1924 Louis de- Schrodinger came up gave the theory Broglie said that with partial radiation just like 1927 Heisenberg differential equation that energy is energy is also radiation has both for the wave fn of said about composed of quantized. wave and particle uncertainty principle. particles. Time quanta Explained PE effect nature evolution of quantum state Department of Physics - MIT, Bangalore 36 Problems: 1. 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 2. 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 3. 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: 70o Department of Physics - MIT, Bangalore 37  4. An electron has a kinetic energy of 3.0 eV. (a) 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 5. The lifetime of an excited atom is given as 1.0 x 10-8 s. Using the uncertainty principle, compute the line width Δf produced by this finite lifetime? Ans: 8.0 x 106 Hz 6. 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 = 97 c, vPROTON = 1.8 x 107 m/s Department of Physics - MIT, Bangalore 38  Practice Questions 1) 2) 3) Department of Physics - MIT, Bangalore 39  Practice Questions 4) 5) Department of Physics - MIT, Bangalore 40 https://www.youtube.com/watch?v=m7gXgHgQGhw Department of Physics - MIT, Bangalore 41