PHY1072 Applied Physics for Engineers PDF

APPLIED PHYSICS FOR ENGINEERS [SUBJECT CODE: PHY1072] I Year Common Course Material for Computer Stream DEPARTMENT OF PHYSICS MANIPAL INSTITUTE OF TECHNOLOGY  Contents 1 LASERS AND FIBRE O...

APPLIED PHYSICS FOR ENGINEERS [SUBJECT CODE: PHY1072] I Year Common Course Material for Computer Stream DEPARTMENT OF PHYSICS MANIPAL INSTITUTE OF TECHNOLOGY  Contents 1 LASERS AND FIBRE OPTICS............................................................... 3 CHARACTERISTICS OF LASERS............................................................................... 3 INTERACTION OF RADIATION WITH MATTER.................................................... 3 ESSENTIAL COMPONENTS OF A LASER.............................................................. 10 He-Ne LASER............................................................................................................... 10 RUBY LASER.............................................................................................................. 11 SEMICONDUCTOR LASER....................................................................................... 12 OPTICAL FIBRES........................................................................................................ 16 ACCEPTANCE ANGLE, ACCEPTANCE CONE AND NUMERICAL APERTURE17 TYPES OF OPTICAL FIBRES.................................................................................... 19 ATTENUATION IN OPTICAL FIBRES..................................................................... 22 QUESTIONS................................................................................................................. 26 PROBLEMS.................................................................................................................. 27 2 QUANTUM PHYSICS........................................................................... 30 BLACKBODY RADIATION AND PLANCK’S HYPOTHESIS............................... 30 PHOTOELECTRIC EFFECT....................................................................................... 33 COMPTON EFFECT.................................................................................................... 36 PHOTONS AND ELECTROMAGNETIC WAVES [DUAL NATURE OF LIGHT]. 39 DE BROGLIE HYPOTHESIS - WAVE PROPERTIES OF PARTICLES.................. 39 THE QUANTUM PARTICLE...................................................................................... 41 DOUBLE–SLIT EXPERIMENT REVISITED............................................................. 44 UNCERTAINTY PRINCIPLE...................................................................................... 44 QUESTIONS................................................................................................................. 45 PROBLEMS.................................................................................................................. 46 3 QUANTUM MECHANICS.................................................................... 50 AN INTERPRETATION OF QUANTUM MECHANICS.......................................... 50 THE SCHRÖDINGER EQUATION............................................................................ 52 PARTICLE IN AN INFINITE POTENTIAL WELL (PARTICLE IN A “BOX”)...... 52 A PARTICLE IN A POTENTIAL WELL OF FINITE HEIGHT................................ 55 TUNNELLING THROUGH A POTENTIAL ENERGY BARRIER........................... 56 THE SIMPLE HARMONIC OSCILLATOR................................................................ 57 1  QUESTIONS................................................................................................................. 58 PROBLEMS.................................................................................................................. 59 4 INTRODUCTION TO QUANTUM COMPUTING.............................. 61 4.1 Mathematical Basis of Quantum Mechanics........................................................... 62 4.2 The Postulates of Quantum Mechanics................................................................... 72 4.3 Quantum Computing............................................................................................... 79 QUESTIONS............................................................................................................... 108 PROBLEMS................................................................................................................ 110 2  1 LASERS AND FIBRE OPTICS OBJECTIVES: To explain basic interactions of radiation with matter. To comprehend the basic principle of lasing system and their applications. To understand the basic working principle of optical fibre and their applications. To classify different types of optical fibres and explain their merits & demerits. CHARACTERISTICS OF LASERS Laser light is highly monochromatic, highly coherent, highly directional and can be sharply focused. Each of these characteristics that are not normally found in ordinary light makes laser a unique and the most powerful tool. Lasers find a wide variety of applications in the field of scientific research, engineering and medicine. INTERACTION OF RADIATION WITH MATTER There are three possible ways through which interaction of radiation and matter can take place, absorption of radiation is one means, and the emission of radiation, as suggested by Einstein in 1917, can occur in two ways namely the spontaneous emission, and the stimulated emission. All the three are described as follows by considering an atom as a system. Absorption: Absorption of a photon of frequency f takes place when the energy difference E2 – E1 of the allowed energy states of the atomic system equals the energy hf of the photon. Then the photon disappears and the atomic system moves to upper energy state E2. Figure 1.1: Absorption 3 Spontaneous Emission: Spontaneous emission is the emission of a photon when a system transits from a higher energy state to a lower energy state without the aid of any external agency. The average life time of the atomic system in the excited state is of the order of 10–8 s. After the life time of the atomic system in the excited state, it comes back to the state of lower energy on its own accord by emitting a photon of energy hf = E2– E1. In an ordinary light source, the radiation of light from different atoms is not coherent. The radiations are emitted in different directions in random manner. Such type of emission of radiation is called spontaneous emission. Figure 1.2: Spontaneous Emission Stimulated Emission: When a photon (called stimulating photon) of suitable frequency interacts with an excited atomic system, the latter comes down to ground state before its life time. Such an emission of radiation is called stimulated emission. In stimulated emission, both the stimulating photon and the stimulated photon are of same frequency, same phase and are in same state of polarization. These photons are emitted in the same direction. In other words, these two photons are coherent. This phenomenon is responsible for laser action. Figure 1.3: Stimulated Emission 4 POPULATION INVERSION From Boltzmann statistics, the ratio between population of atoms in two energy states E1 and E2 at equilibrium temperature T is, 𝑵𝟐 = 𝒆−(𝑬𝟐 −𝑬𝟏 )/𝒌𝑻 𝑵𝟏 1.1 where k is Boltzmann constant, N1 is density of atoms with energy E1 & N2 is density of atoms with energy E2. Under normal condition, where populations are determined only by the action of thermal agitation, population of the atoms in upper energy state is less than that in lower energy state (i.e. N2 < N1, Figure 1.4 a). Figure 1.4: (a) Normal thermal equilibrium distribution of atomic systems (b) An inverted population. For the stimulated emission rate to exceed the absorption rate it is necessary to have higher population of upper energy state than that of lower energy state. This condition is called population inversion (N2 > N1, Figure 1.4b). This is a non-equilibrium condition and is facilitated by the presence of energy states called ‘metastable states’ where the average life time of the atom is 10-3 s which is much longer than that of the ordinary excited state ( 10-8s). EINSTEIN’S COEFFICIENTS Einstein coefficients give the probability associated with the absorption and emission processes. Consider two energy states 𝐸1 and 𝐸2 of a system of atoms (E2 > E1 ). Let N₁ and N2 be the density of atoms with energies 𝐸1 and E2 respectively, per unit volume of the system. Let radiations with a continuous spectrum of frequencies be incident upon the system. Let there be a radiation of frequency 𝑓 such that 𝑓 = (𝐸2 − 𝐸1 )/ℎ. Also let I𝑓 𝑑𝑓 be the energy incident/unit volume of the system considering only those radiations whose frequencies lie in the range 𝑓 and 𝑓 +𝑑𝑓. Then, 𝐼𝑓 represents the energy density of frequency 𝑓. Let us consider the absorption, and the two emission processes case by case. 5 (1) In the case of absorption, an atom in the level 𝐸₁ can go to the level 𝐸2 , when it absorbs a radiation of frequency 𝑓 such that 𝑓 = (𝐸2 − 𝐸1 )/ℎ (Fig. 1.1). The number of such absorptions per unit time, per unit volume is called rate of absorption. The rate of absorption depends upon, (a) the number density of lower energy state, i.e., 𝑁₁, and, (b) the energy density i.e., 𝐼𝑓. ∴ Rate of absorption ∝ 𝑵𝟏 𝑰𝒇 Rate of absorption = 𝑩₁₂ 𝑵₁ 𝑰𝒇 1.2 where, 𝐵₁₂ is the constant of proportionality called Einstein coefficient of induced absorption. (2) In the case of spontaneous emission, an atom in the higher energy level 𝐸2 undergoes transition to the lower energy level 𝐸1 , voluntarily by emitting a photon (Fig. 1.2). Since it is a voluntary transition, it is independent of the energy density of any frequency in the incident radiation. The number of such spontaneous emissions per unit time, per unit volume is called rate of spontaneous emission which is proportional to only the number density in the higher energy state, i.e., 𝑁2. ∴ Rate of spontaneous emission = 𝑨𝟐𝟏 𝑵₂ 1.3 where, 𝐴21 is the constant of proportionality called the Einstein coefficient of spontaneous emission. (3) In case of stimulated emission, the system requires an external photon of appropriate frequency = [(𝐸₂ − 𝐸₁)/ℎ], to stimulate the atom for the corresponding downward transition, and thereby cause emission of stimulated photons (Fig. 1.3), the energy density 𝐼𝑓 , has a role to play in this case. The number of stimulated emissions per unit time per unit volume, called rate of stimulated emission, is proportional to, a) the number density of the higher energy state, i.e., 𝑁₂ and, b) the energy density, i.e., 𝐼𝑓. ∴ Rate of stimulated emission ∝ 𝑵𝟐 𝑰𝒇 Rate of stimulated emission = 𝑩𝟐𝟏 𝑵𝟐 𝑰𝒇 1.4 6 where, 𝐵21 is the constant of proportionality called the Einstein coefficient of simulated emission. In thermal equilibrium, the total energy of the system remains unchanged in spite of the interaction that is taking place between itself and the incident radiation. Under such a condition, the number of photons absorbed by the system per second must be equal to the number of photons it emits per second by both the stimulated and the spontaneous emission processes. ∴ At thermal equilibrium, Rate of absorption = Rate of spontaneous emission + Rate of stimulated emission, ∴ From Eqs (1.1), (1.2) & (1.3), we have 𝑩₁₂ 𝑵₁ 𝑰𝒇 = 𝑨𝟐𝟏 𝐍𝟐 + 𝑩𝟐𝟏 𝑵𝟐 𝑰𝒇 1.5 Or 𝐼𝑓 (𝐵₁₂ 𝑁₁ − 𝐵21 𝑁2 ) = 𝐴21 N2 𝐴21 N2 Or, 𝐼𝑓 = (𝐵₁₂ 𝑁₁ − 𝐵21 𝑁2 ) By rearranging the above equation, we get, 𝑨 𝟏 𝑰𝒇 = 𝑩𝟐𝟏 [ 𝑩₁₂ 𝑵₁ ] 𝟐𝟏 −𝟏 𝑩𝟐𝟏 𝐍𝟐 1.6 But, by Boltzmann's law, we have, 𝒉𝒇 𝒉𝒇 𝑵𝟐 𝑵 = 𝒆−(𝑬𝟐 −𝑬𝟏 )/𝒌𝑻 = 𝒆−𝒌𝑻 ∴ 𝑵𝟏 = 𝒆𝒌𝑻 𝑵𝟏 𝟐 1.7 Equation (1.6) becomes, 𝑨 𝟏 𝑰𝒇 = 𝑩𝟐𝟏 [ 𝒉𝒇 ] 1.8 𝟐𝟏 𝑩₁₂ 𝒆𝒌𝑻 − 𝟏 𝑩𝟐𝟏 According to Planck's law, the equation for 𝐼𝑓 , is, 𝟖𝝅𝒉𝒇𝟑 𝟏 𝑰𝒇 = [ 𝒉𝒇 ] 𝒄𝟑 𝒆𝒌𝑻 − 𝟏 1.9 Now, comparing the equations (1.8) and (1.9), term by term on the basis of positional identity, we have, 7 𝐴21 8𝜋ℎ𝑓 3 𝐵₁₂ = and, =1 or 𝐵12 = 𝐵21 𝐵21 𝑐3 𝐵21 Hence the probability of induced absorption is equal to the probability of stimulated emission. ℎ𝑓 𝐴21 Hence equation (1.8) can be rearranged as = 𝐼𝑓 (𝑒 𝑘𝑇 − 1) 1.10 𝐵21 Because of the above identity, the subscripts could be dropped, and 𝐴21 & 𝐵21 can be represented simply as A and B and Eq (1.8) can be rewritten. 𝐀 ∴ At thermal equilibrium the equation for energy density is, 𝑰𝒇 = 𝒉𝒇 𝑩[𝒆𝒌𝑻 − 𝟏] 1.11 INTERPRETATION OF EINSTEIN’S COEFFICIENTS: Dependence of nature of emission on frequency: 𝐴21 8𝜋ℎ𝑓 3 𝐴21 We have, = 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 ∝ 𝑓3 𝐵21 𝑐3 𝐵21 Since 𝑓 = (Δ𝐸/ℎ), the above proportionality implies that for large f values. For higher Δ𝐸 values, the transitions in the system are such that, the probability of spontaneous emission supersedes that for stimulated emissions by a large margin. Case (i) 𝒉𝒇 >> 𝒌𝑻 : ℎ𝑓 When the frequency of radiation is high, ℎ𝑓 >> 𝑘𝑇, due to which 𝑒 𝑘𝑇 ≫ 1 𝐴21 Hence in Equation (1.10), ≫ 1, i.e.. A21≫ 𝐵21 𝐵21 Hence the spontaneous emissions will be much larger in number than the stimulated emissions. Case (ii) ≈ 𝒌𝑻 : ℎ𝑓 For ℎ𝑓 ≈ 𝑘𝑇, 𝑒 𝑘𝑇 will be low and comparable to 1. ∴ 𝐴21 & 𝐵21 become comparable, which means, the process of stimulated emissions become significant. Case (iii) >1, then the number of successfully propagated modes is. 2 Single mode step index optical fibre: A single mode optical fibre consists of a core having a uniform refractive index n1 that abruptly decreases at the core-cladding interface to a lower value n2, the refractive index of the cladding. The diameter of the core is narrow (5-10m) generally a few times the wavelength of the light propagating through it. Only rays nearly parallel to the fibre axis will travel through. It supports a single mode propagation because of its narrow core (Figure 1.18). Figure 1.18: Single mode step-index fibre Step-Index Multimode fibre: In this case also the refractive index profile is similar to step index fibre i.e., fibre consists of a core having a uniform refractive index n1 that abruptly 20 decreases at the core-cladding interface to a lower value n2 the refractive index of the cladding. But the diameter of the core is large (50-200 m). The comparatively large central core makes it rugged and easily infused with light, as well as easily terminated and coupled. It also supports a large number of modes for propagation (Figure 1.19). Figure 1.19: Multi mode step-index fibre Graded-Index Multimode fibre (GRIN): It consists of a core whose refractive index decreases gradually from its axis radially outward and becomes equal to the refractive index of the cladding at the core-cladding interface. The refractive index of the cladding remains uniform. Dimensions of the core and cladding are similar to that of step index multimode fibres. It supports a large number of modes for propagation because of its large core diameter (Figure 1.20). 21  Figure 1.20: Multi mode graded index fibre ATTENUATION IN OPTICAL FIBRES Attenuation is the loss of power of the light signal that occurs during its propagation through the optical fibre. The main sources of attenuation are 1. absorption 2. scattering 3. other losses Absorption: Absorption of light during propagation occurs due to the impurities present in the fibre material and also due to the intrinsic nature of the material itself. The impurities generally present are a) Transition metals such as iron, chromium, cobalt, copper etc. b) The hydroxy ions (OH-) that enter into the fibre material at the time of fabrication. The photons absorbed by the impurities may be lost as heat or may be reemitted as light energy of different wavelengths and different phase from the one that is propagated. Hence it results in a loss. Intrinsic absorption occurs by the pure material itself even if the material is free from impurities and inhomogeneities. Intrinsic absorption though quite less compared to the loss due to the impurities, it cannot be eliminated. Scattering: Glass is a heterogeneous mixture of oxides of silicon, phosphorus, germanium etc. Structural inhomogeneities in the core index will set in the fibre material during solidification of glass from its molten state. It will also result in a fluctuation of the molecular density. These inhomogeneities act as scattering centers. Since their dimensions are smaller than  the wavelength of the light propagated through the fibre, the energy loss that occurs due to such 1 scattering resemble Rayleigh scattering that is proportional to. The losses due to these 4 scattering cannot be eliminated by any process. There are other structural inhomogeneities & 22 defects that set in during fabrication of the fibre that contribute to the loss due to scattering. Their sources are trapped gas bubbles, unreacted starting materials etc. However, these can be reduced to a great extent by improved methods of manufacturing. Other losses: a) Due to dimensional irregularities and imperfections in the fibres (that are called microscopic bends) (Figure 1.21) the light may not sustain total internal reflection. The energy will escape from the core. Figure 1.21: Signal attenuation in optical fibre due to microscopic bending b) Macroscopic bends occur during wrapping the fibre on a spool or negotiating a curve during cable laying. Fibres can withstand bends of curvature up to about 10cm without significant loss. For higher curvature (smaller radius of curvature) than this, the loss increases exponentially. Amplification is therefore needed in communication applications at regular intervals in order to compensate for the losses that occur despite all precautions. An optical repeater is used to boost the signal. DISTORTION IN OPTICAL FIBRES A light pulse launched into a fibre decreases in amplitude, as it travels along the fibre, due to losses in the fibre. It also spreads during its travel. The pulse received at the output is wider than input pulse, as shown in Figure 1.22. Such distortion arises due to dispersion effects. Figure 1.22: Broadening of the signal due to distortion 23 There are three mechanisms which contribute to the distortion (signal spreading) of the light pulse in a fibre. (i) Material dispersion: The material dispersion is due to the dependence of the refractive indices of glass and consequently the group velocity of the optical signal. A LED emits a broad spectrum of light. Whenever a LED is used as a source, the broad spectrum contains a number of wavelengths. These different wavelengths of light are travelling through optical fibre at different speeds because the refractive indices of the glass vary with wavelengths. The short wavelength waves travel slower than long wavelength waves in a material. Hence, narrow pulses of light tend to broaden as they travel down the optical fibre. This is known as material dispersion (Fig. 1.23 (a)). Material dispersion is present in both single mode and multimode fibres. The material dispersion is given by the equation λ (Δλ) d2 𝑛 D𝑚 = L 1.20 c dλ2 where λ is the peak wavelength, Δλ is the spectral width, L is the length of the core, and n is the refractive index of the core. To cite an example, an LED operating at 820 nm and having a spectral width of 38 nm results in a dispersion of about 3 ns/km in a certain fibre. In the same fiber, dispersion can be reduced to 0.3 ns/km using a laser diode operating at 1140 nm and having a 3 nm spectral width. Thus, material dispersion can be reduced using a more and more monochromatic source. Figure 1.23: (a) Material dispersion (b) Waveguide dispersion (c) Intermodal dispersion (ii) Waveguide dispersion: Waveguide dispersion arises from the guiding properties of the fibre. Pulses at different wavelengths, through propagating in the same mode, travel at slightly 24 different angles. This is known as waveguide dispersion (Fig. 1.23 (b)). It is present in both single mode and multimode fibres (iii) Intermodal dispersion: A ray of light launched into a fibre follows different zigzag paths. When numerous modes are propagating in a fibre, they travel with different net velocities with respect to fibre axis. Parts of the wave arrive at the output before other parts leading to a spread of the input pulse. This is known as intermodal dispersion (Fig. 1.23 (c)). It does not depend on the spectral width of the source. A light pulse from an ideal monochromatic source (Δλ =0) would still exhibit spreading. In single- mode fibres, since there in only one mode, there is no intermodal dispersion. APPLICATIONS OF OPTICAL FIBRES i) Fibre Optic Networking: A Fibre optic network is a telecommunication network with optical fibre as the primary transmission medium. It is designed in such a way that it makes full use of the unique features of optical fibres. Over the years, optical fibre has become the preferred medium for provision of the major infrastructure for voice, video and data transmission, because it offers far greater bandwidth and is less bulky than copper cables. Figure (1.24) shows the structure of a simple optical network. It consists of optical nodes interconnected with optical fibre links. An optical node is a multifunctional element which basically acts as a transceiver unit capable of receiving, transmitting and processing the optical signal. A signal carried on a dedicated wavelength from source to a destination node is known as a light path (i.e. λ1 and λ2 are two light paths in this particular case). The data can be sent over these light paths once the connections are set up. Figure 1.24: Structure of Fibre Optic Network 25 ii) Fibre Optic Communication: Optical communication is communication at a distance using light to carry information. An optical communication system uses a transmitter, which encodes a message into an optical signal, a channel (optical fibre), which carries the signal to its destination, and a receiver, which reproduces the message from the received optical signal (Fig.1.25). Optical communication has several advantages – can carry large data in digital form, interference and noise free. Figure 1.25: Simplified block diagram of optical communication system Optical fibres are also used in sensors, flexible fibrescope (endoscope) and other industrial applications. QUESTIONS 1.1 LASER 1 Mention the characteristics of a laser beam. 2 Explain the following terms with reference to lasers: (a) spontaneous emission (b) stimulated emission (c) metastable state (d) population inversion (e) pumping (f) active medium (g) resonant cavity. 3 Explain the principle of a laser. 4 Explain construction and operation of ruby laser with necessary diagrams. 5 Explain construction and operation of He-Ne laser with necessary diagrams. 6 Write a note on semiconductor lasers. 26  7 Mention any four applications of laser. 8 Explain LASER Cooling and its application 1.2 OPTICAL FIBRES 11 What is a critical angle? Define total internal reflection 12 With necessary diagram, derive an expression for angle of acceptance and numerical aperture. 13 Define fractional refractive index change  and get the relation between  & numerical aperture NA. 14 What is skip distance? With neat diagram, derive an expression for it. 15 What are different types of optical fibres? Briefly explain them with diagrams. 16 What are the different types of attenuations possible in optical fibres? Explain them. Explain different types of distortion in optical fibres. 17 Write a note on Fiber optic networking 18 Explain Fiber optic communication PROBLEMS 1.1 LASER 1 A three level laser of the type shown in figure, emits laser light at a wavelength 550 nm, near the centre of the visible band. If the optical mechanism is shut off, what will be the ratio of the population of the upper level E2 to that of the lower level E1 at 300 K ? At what temperature for the condition of (a) would the ratio of populations be half ? Ans: (a) 1.77 x 10-38 (b) 37800 K 2 A pulsed ruby laser has a ruby rod (Al2O3 doped with Cr2O3) as an active medium, which is 6 cm long and 1 cm in diameter. There is one chromium ion (active centre, with energy levels of the type shown in the figure) for every 3500 aluminium ions. The ruby laser light has a wavelength of 694.4 nm. Suppose that all the chromium ions are in metastable state (E2) and none are in ground state (E1). How much energy is there in a single laser pulse if all 27  these ions come down to ground state in a single stimulated emission chain reaction episode ? Density of Al2O3 is 3700 kg/m3. Molar mass of Al2O3 is 0.102 kg/mol. Ans: Total energy per pulse= 16.9 J 3 A ruby laser emits light at a wavelength of 694.4 nm. If a laser pulse is emitted for 12.0 ps and the energy release per pulse is 150 mJ, (a) what is the length of the pulse, and (b) how many photons are there in each pulse ? (a) 3.6 x 10-3 m (b) 5.25 x 1017 4 It is entirely possible that techniques for modulating the frequency or amplitude of a laser beam will be developed so that such a beam can serve as a carrier for television signals, much as microwave beams do now. Assume also that laser systems will be available whose wavelengths can be precisely tuned to anywhere in the visible range (400 nm to 700 nm). If a television channel occupies a bandwidth of 10 MHz, how many channels could be accommodated with this laser technology ? Comment on the intrinsic superiority of visible light to microwaves as carriers of information. Ans: 3.27 x 107 5 A He-Ne laser emits light at a wavelength of 632.8 nm and has an output power of 2.3 mW. How many photons are emitted each minute by this laser when operating ? Ans: 4.4 x 1017 /min 6 An atom has two energy levels with a transition wavelength of 582 nm. At 300 K, 4.0 x 1020 atoms are in the lower state. (a) How many occupy the upper state under conditions of thermal equilibrium ? (b) Suppose, instead, that 7.0 x 1020 atoms are pumped into upper state, with 4.0 x 1020 atoms in the lower state. How much energy could be released in a single laser pulse ? Ans: (a) 6.6 x 10-16  0 (b) 240 J 7 Calculate the ratio of, (i) Einstein’s coefficients (ii) rate of stimulated to spontaneous emissions, for a system at 300 K in which radiations of wavelength 1.39 µm are emitted. [Ans: The ratio of [A21/B12] is 6.2 x 10-15, and ratio of rate of stimulated to spontaneous emissions 10-15] 28 1.2 OPTICAL FIBRE 8 A step index optical fibre 63.5 m in core-diameter has a core of refractive index 1.53 and a cladding of index 1.39. Determine (a) the numerical aperture for the fibre, (b) the critical angle for core-cladding interface, (c) the acceptance cone half-angle (the maximum entrance angle) (d) the number of reflections in 1.0 m length of the fibre for a ray at the maximum entrance angle, (e) the number of reflections in 1.0 m length of the fibre for a ray at half the maximum entrance angle. Ans: a) 0.639 b) 65.29° (c) 39.71° (d) 7.25 x 103 (e) 358 x 103 9 A glass optical fibre of refractive index 1.450 is to be clad with another to ensure total internal reflection that will contain light traveling within 5° of the fibre-axis. What maximum index of refraction is allowed for the cladding? Ans: 1.444 10 The numerical aperture of an optical fibre is 0.2 when surrounded by air. Determine the refractive index of its core. The refractive index of the cladding is 1.59. Also find the acceptance cone half-angle when the fibre in water. Refractive index of water is 1.33. Ans: 1.6, 8.65° 11 The angle of acceptance of an optical fibre is 30° when kept in air. Find the angle of acceptance when it is in a medium of refractive index 1.33. Ans: 22° 29  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. BLACKBODY RADIATION AND PLANCK’S HYPOTHESIS A black body is 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. The radiation is absorbed in the walls of the cavity. This causes a heating of the cavity walls. The oscillators in the cavity walls vibrate and re-radiate at wavelengths corresponding to the temperature of the cavity, thereby producing standing waves. 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. Figure 2-1: A physical model of a blackbody. The black body is an ideal absorber of incident radiation. A black-body reaches thermal equilibrium with the surroundings when the incident radiation power is balanced by the power re-radiated. The emitted "thermal" radiation from a black body characterizes the equilibrium temperature of the black-body. The nature of radiation from a blackbody does not depend on the material of which the walls are made. Basic laws of radiation (1) All objects emit radiant energy. 30 (2) Hotter objects emit more energy (per unit area) than colder objects. The total power of the emitted radiation is proportional to the fourth power of temperature. This is called Stefan’s Law and is given by, P =  A e T4 2-1 where P is power radiated from the surface of the object (W), T is equilibrium surface temperature (K), σ is Stefan-Boltzmann constant (= 5.670 x 10−8 W/m2K4 ), A is surface area of the object (m2) and e is emissivity of the surface (e =1 for a perfect blackbody). (3) The peak of the wavelength distribution shifts to shorter wavelengths as the black body temperature increases. This is Wien’s Displacement Law and is given by, λm T = constant = 2.898 × 10−3 m.K , or λm  T−1 2-2 where λm is the wavelength corresponding to peak intensity and T is equilibrium temperature of the blackbody. Figure 2-2: Intensity of blackbody radiation versus wavelength at two temperatures. (4) Rayleigh-Jeans Law: This law tries to explain the distribution of energy from a black body. 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 ) = 2-3 4 kB is Boltzmann’s constant, c is speed of light in vacuum, T is equilibrium blackbody temperature. 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. 31  Figure 2-3: Comparison of experimental results and the curve predicted by the Rayleigh–Jeans law for the distribution of blackbody radiation 5) Planck‘s Law: Max Planck developed a theory of blackbody radiation that leads to an equation for I (,T) that is in complete agreement with experimental results. To derive the law, Planck made two assumptions concerning the nature of the oscillators in the cavity walls: (i) The energy of an oscillator is quantized hence it can have only certain discrete values: En = n h f 2-4 where n is a positive integer called a quantum number, f is the frequency of cavity oscillators, and h is a constant called Planck’s constant. Each discrete energy value corresponds to a different quantum state, represented by the quantum number n. (ii) The oscillators emit or absorb energy only when making a transition from one quantum state to another. Difference in energy will be integral multiples of hf. Figure 2.4: Allowed energy levels for an oscillator with frequency f 32 Planck’s law explains the distribution of energy from a black body which is given by, 2 h c 2 1 2-5 I(  ,T ) =  5 hc λkB T e −1 where I (,T) d is the intensity or power per unit area emitted in the wavelength interval d from a blackbody, h is Planck’s constant, kB is Boltzmann's constant, c is speed of light in vacuum and T is equilibrium temperature of blackbody. The Planck‘s Law gives a distribution that peaks at a certain wavelength, the peak shifts to shorter wavelengths for higher temperatures, and the area under the curve grows rapidly with increasing temperature. This law is in agreement with the experimental data. 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. hc ▪ For very large λ, exp( hckT ) − 1   kT  I(  ,T ) → 2  c −4 k T i.e. I (λ, T) → 0 as λ → . From a fit between Planck's law and experimental data, Planck’s constant was derived to be h = 6.626 × 10–34 J-s. 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 Figure 2.5: (a) Apparatus (b) circuit for studying Photoelectric Effect (T – Evacuated glass/ quartz tube, E – Emitter Plate / Photosensitive material / Cathode, C – Collector Plate / Anode, V – Voltmeter, A - Ammeter) Experimental Observations: 33 Figure 2.6: Photoelectric current versus applied potential difference for two light intensities 1. When plate E is illuminated by light of suitable frequency, electrons are emitted from E and a current is detected in A (Figure 2.5). 2. Photocurrent produced vs potential difference graph shows that kinetic energy of the most energetic photoelectrons is, Kmax = e Vs 2-6 where Vs is stopping potential 3. Kinetic energy of the most energetic photoelectrons is independent of light intensity. 4. Electrons are emitted from the surface of the emitter almost instantaneously 5. No electrons are emitted if the incident light frequency falls below a cutoff frequency. 6. Kinetic energy of the most energetic photoelectrons increases with increasing light frequency. Classical Predictions: 1. If light is really a wave, it was thought that if one shine of light of any fixed wavelength, at sufficient intensity on the emitter surface, electrons should absorb energy continuously from the em waves and electrons should be ejected. 2. As the intensity of light is increased (made it brighter and hence classically, a more energetic wave), kinetic energy of the emitted electrons should increase. 3. Measurable / larger time interval between incidence of light and ejection of photoelectrons. 4. Ejection of photoelectron should not depend on light frequency 5. In short experimental results contradict classical predictions. 6. Photoelectron kinetic energy should not depend upon the frequency of the incident light. Einstein’s Interpretation of electromagnetic radiation: 34 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 theory of photoelectric effect: A photon of the incident light gives all its energy hf to a single electron (absorption of energy by the electrons is not a continuous process as envisioned in the wave model) and the kinetic energy of the most energetic photoelectron, Kmax = hf −  (Einstein’s photoelectric equation) 2-7  is called the work function of the metal. It is the minimum energy with which an electron is bound in the metal. All the observed features of photoelectric effect could be explained by Einstein’s photoelectric equation: 1. Equation shows that Kmax depends only on frequency of the incident light. 2. Almost instantaneous emission of photoelectrons due to one -to –one interaction between photons and electrons. 3. Ejection of electrons depends on light frequency since photons should have energy greater than the work function  in order to eject an electron. 4. The cutoff frequency fc is related to  by fc =  /h. If the incident frequency f is less than fc , there is no emission of photoelectrons. The graph of kinetic energy of the most energetic photoelectron Kmax vs frequency f is a straight line, according to Einstein’s equation. 35 Figure 2.7: A representative plot of Kmax versus frequency of incident light for three different metals 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. This scattering phenomenon is known as Compton Effect. Classical Predictions: Oscillating electromagnetic waves (classically, X-rays are em waves) incident on electrons should have two effects: i) oscillating electromagnetic field causes oscillations in electrons. Each electron first absorbs radiation as a moving particle and then re- radiates in all directions as a moving particle and thereby exhibiting two Doppler shifts in the frequency of radiation. ii) radiation pressure should cause the electrons to accelerate in the direction of propagation of the waves. Because different electrons will move at different speeds after the interaction, depending on the amount of energy absorbed from electromagnetic waves, the scattered waves at a given angle will have all frequencies (Doppler- shifted values). Compton’s experiment and observation: Compton measured the intensity of scattered X- rays from a solid target (graphite) as a function of wavelength for different angles. The experimental setup is shown in Figure 2.8. Contrary to the classical prediction, only one frequency for scattered radiation was seen at a given angle. This is shown in the Figure 2.9. The graphs for three nonzero angles show two peaks, one at o and the other at ’ >o. The shifted peak at ’ is caused by the scattering of X-rays from free electrons. Shift in wavelength was predicted by Compton to depend on scattering angle as 𝒉 2-8 𝝀′ − 𝝀 = 𝒎𝒄 (𝟏−𝒄𝒐𝒔 𝜽) where m is the mass of the electron, c is velocity of light, h is Planck’s constant. ℎ This is known as Compton shift equation, and the factor 𝑚𝑐 is called the Compton ℎ wavelength and 𝑚𝑐 = 2.43 pm for electron. 36  Figure 2.8 Schematic diagram of Compton’s apparatus. The wavelength is measured with a rotating crystal spectrometer for various scattering angles θ. Figure 2.9: Scattered x-ray intensity versus wavelength for Compton scattering at  = 0°, 45°, 90°, and 135° showing single frequency at a given angle Derivation of the Compton shift equation: Compton could explain the experimental result by treating the X-rays not as waves but rather as point like particles (photons) having energy E = hfo = hc/o , momentum p = hf/c = h/ and zero rest energy. Photons collide elastically with free electrons initially at rest and moving relativistically after collision. Let o , po = h/o and Eo = hc/o be the wavelength, momentum and energy of the incident photon respectively. ’, p’ = h/’ and E’ = hc/’ be the corresponding quantities for the scattered photon. We know that, for the electron, the total relativistic energy 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4 Kinetic energy K = E − m c2 1 And momentum p =  mv. where = 1 − vc 2 2 v and m are the speed and mass of the electron respectively. 37  Figure 2.10: Quantum model for X-ray scattering from an electron In the scattering process, the total energy and total linear momentum of the system must be conserved. For conservation of energy we must have, Eo = E’ + K ie, Eo = E’ + (E − m c2) Or Eo − E’ + m c2 = 𝐸 = √𝑝2 𝑐 2 + 𝑚2 𝑐 4 Squaring both the sides, (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 + 𝑚2 𝑐 4 = 𝑝2 𝑐 2 + 𝑚2 𝑐 4 For conservation of momentum, x-component: 𝑝𝑜 = 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝 𝑐𝑜𝑠 𝜙 y-component: 0 = 𝑝′ 𝑠𝑖𝑛 𝜃 − 𝑝 𝑠𝑖𝑛 𝜙 Rewriting these two equations 𝑝𝑜 − 𝑝′ 𝑐𝑜𝑠 𝜃 = 𝑝 𝑐𝑜𝑠 𝜙 𝑝′ 𝑠𝑖𝑛 𝜃 = 𝑝 𝑠𝑖𝑛 𝜙 Squaring both the sides and adding, 𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 = 𝑝2 Substituting this 𝑝2 in the equation : (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = 𝑝2 𝑐 2 , one gets (𝐸𝑜 − 𝐸′)2 + 2(𝐸𝑜 − 𝐸′) 𝑚𝑐 2 = (𝑝𝑜2 − 2𝑝𝑜 𝑝′ 𝑐𝑜𝑠 𝜃 + 𝑝′2 )𝑐 2 Substituting photon energies and photon momenta one gets ℎ𝑐 ℎ𝑐 2 ℎ𝑐 ℎ𝑐 ℎ𝑐 2 ℎ𝑐 ℎ𝑐 ℎ𝑐 2 ( 𝜆𝑜 − 𝜆′ ) +2( 𝜆𝑜 − 𝜆′ ) 𝑚𝑐 2 = ( 𝜆𝑜 ) − 2( 𝜆𝑜 ) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 + ( 𝜆′ ) Simplifying one gets 38  ℎ𝑐 2 ℎ𝑐 ℎ𝑐 ℎ𝑐 2 1 1 ℎ𝑐 2 ℎ𝑐 ℎ𝑐 ( 𝜆 ) − 2 ( 𝜆 ) ( 𝜆′ ) + ( 𝜆′ ) + 2 ℎ𝑐 ( 𝜆𝑜 − 𝜆′ ) 𝑚𝑐 2 = ( 𝜆𝑜 ) − 2( 𝜆𝑜 ) ( 𝜆′ ) 𝑐𝑜𝑠 𝜃 𝑜 𝑜 ℎ𝑐 2 + ( 𝜆′ ) ℎ𝑐 ℎ𝑐 i.e., − + ( 𝜆1𝑜 − 1 𝜆′ ) 𝑚𝑐 2 = − 𝑐𝑜𝑠 𝜃 𝜆𝑜 𝜆′ 𝜆𝑜 𝜆′ ′ ℎ𝑐 OR, (𝜆𝜆−𝜆𝜆𝑜′ ) 𝑚𝑐 2 = (1 − 𝑐𝑜𝑠 𝜃) 𝑜 𝜆𝑜 𝜆′ Compton shift: 𝒉 𝝀′ − 𝝀𝒐 = (𝟏 − 𝒄𝒐𝒔 𝜽) 𝒎𝒄 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. DE BROGLIE HYPOTHESIS - WAVE PROPERTIES OF PARTICLES We have seen that light comes in discrete units (photons) with particle properties (energy E and momentum p) that are related to the wave-like properties of frequency and wavelength. 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. 𝒉 𝒉 2-9 de Broglie wavelength: 𝝀 = = 𝒑 𝒎𝒗 where h is Planck’s constant and p is momentum of the quantum particle, m is mass of the particle, and v is speed of the particle. The electron accelerated through a potential difference of V, has a non-relativistic kinetic energy 1 2 𝑚 𝑣 2 = 𝑒 ∆𝑉 where e is electron charge. Hence, the momentum (p) of an electron accelerated through a potential difference of V is 𝒑 = 𝒎 𝒗 = √𝟐 𝒎 𝒆 ∆𝑽 2-10 39  𝐸 Frequency of the matter wave associated with the particle is , where E is total relativistic ℎ energy of the particle Davisson and Germer experiment This is an experiment on diffraction of accelerated electrons by crystals to establish the wave nature of electrons. The regular spacing of the atoms in a crystal act as a grating for an electron beam producing a diffraction pattern by electron scattering. The result of this experiment resembles the result of the x ray diffraction giving experimental evidence for de Broglie hypothesis. Figure 2.11: Schematic of Davisson-Germer experiment (left). Polar plot of intensities after scattering for incident electron energy of 54 eV. (right). A beam of electrons from a heated filament accelerated to a potential V is collimated and allowed to strike a single crystal of nickel. Electrons are scattered in all directions by the atoms in the crystal. The intensity of the scattered electron beam is measured by a detector which can be moved to any angle 𝜙 relative to the incident beam and is shown in the polar plot of  verses the intensity ( r, the radius vector). The most intense reflection of electron beam at an angle 𝜙 = 50° , for an accelerating potential 54 V is observed. An electron of mass m accelerated to a potential V has 1 kinetic energy, 𝑚𝑣 2 = 𝑒 𝑉 where , v = velocity. 2 And the momentum, 𝑝 = 𝑚 𝑣 = √2 𝑚 𝑒 𝑉 ℎ ℎ ℎ  de Broglie wavelength, 𝜆 = 𝑝 = 𝑚𝑣 = √2 𝑚 𝑒 𝑉 40 Substituting V = 54 V, we get the value for the de Broglie wavelength as  = 0.167nm. Wavelength can also be obtained, in this experiment, using Bragg’s law of diffraction. According to which the wavelength is given by the following relationship: 𝑑 𝑠𝑖𝑛𝜙 = 𝑛𝜆 Substituting n = 1 (for the first order diffraction maximum), 𝜙 = 50° (scattering angle corresponding to the first order diffraction maximum), 𝑑 = 0.215 𝑛𝑚 (inter-atomic spacing in nickel), we get  = 0.165nm The values of the wavelength from both the approaches agree very well. This verifies de Broglie’s hypothesis and establishes the wave nature of electrons. Davisson-Germer experiment and G P Thomson’s electron diffraction experiment confirmed de Broglie relationship p = h /. Subsequently it was found that atomic beams, and beams of neutrons, also exhibit diffraction when reflected from regular crystals. Thus de Broglie's formula seems to apply to any kind of matter. Now the dual nature of matter and radiation is an accepted fact and it is stated in the principle of complementarity. This states that wave and particle models of either matter or radiation complement each other. THE QUANTUM PARTICLE Quantum particle is a model by which particles having dual nature are represented. We must choose one appropriate behavior for the quantum particle (particle or wave) in order to understand a particular behavior. 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) as shown in Figure 2.11. Figure 2.11: Section of an ideal wave of single frequency Now to build a localized entity from an infinitely long wave, waves of same amplitude, but slightly different frequencies are superposed (Figure 2.12). 41  Figure 2.12: Superposition of two waves Wave 1 and Wave 2 If we add up large number of waves in a similar way, the small localized region of space where constructive interference takes place is called a wave packet, which represents a quantum particle (Figure 2.13). Figure 2.13: Wave packet Mathematical representation of a wave packet: Superposition of two waves of equal amplitude, but with slightly different frequencies, f1 and f2, traveling in the same direction are considered. The waves are written as 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 𝛥𝜔 𝑘1 +𝑘2 𝜔1 +𝜔2 𝑦 = 2𝐴 [𝑐𝑜𝑠 ( 2 𝑥 − 2 𝑡) 𝑐𝑜𝑠 ( 2 𝑥 − 2 𝑡)] where k = k1 – k2 and  = 1 – 2. Figure 2.14: Beat pattern due to superposition of wave trains y1 and y2 42 The resulting wave oscillates with the average frequency, and its amplitude envelope (in square brackets, shown by the blue dotted curve in Figure 2.14) varies according to the difference frequency. A realistic wave (one of finite extent in space) is characterized by two different speeds. The phase speed, the speed with which wave crest of individual wave moves, is given by 𝜔 𝑣𝑝 = 𝑓 𝜆 or 𝑣𝑝 = (2.11) 𝑘 The envelope of group of waves can travel through space with a different speed than the individual waves. This speed is called the group speed or the speed of the wave packet which is given by (𝛥𝜔 2 ) 𝛥𝜔 𝑣𝑔 = = (2.12) (𝛥𝑘 2 ) 𝛥𝑘 For a superposition of large number of waves to form a wave packet, this ratio is 𝑣𝑔 = 𝑑𝜔 𝑑𝑘 In general these two speeds are not the same. 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 ( ) (2.13) 𝑑𝜆 Relation between group speed (vg) and particle speed (u): 𝐸 2𝜋 2𝜋 2𝜋𝑝 𝜔 = 2𝜋𝑓 = 2𝜋 and 𝑘 = = = ℎ 𝜆 ℎ⁄𝑝 ℎ 2𝜋 𝑑𝜔 𝑑𝐸 𝑑𝐸 ℎ 𝑣𝑔 = = 2𝜋 = 𝑑𝑘 ℎ 𝑑𝑝 𝑑𝑝 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 (2.14) 𝑑𝑘 𝑑𝑝 i.e., we should identify the group speed with the particle speed, speed with which the energy moves. To represent a realistic wave packet, confined to a finite region in space, we need the superposition of large number of harmonic waves with a range of k-values. 43 DOUBLE–SLIT EXPERIMENT REVISITED One way to confirm our ideas about the electron’s wave–particle duality is through an experiment in which electrons are fired at a double slit. Consider a parallel beam of mono- energetic electrons incident on a double slit as in Figure 3.15. Let’s assume the slit widths are small compared with the electron wavelength so that diffraction effects are negligible. An electron detector screen (acts like the “viewing screen” of Young’s double-slit experiment) is positioned far from the slits at a distance much greater than d, the separation distance of the slits. If the detector screen collects electrons for a long enough time, we find a typical wave interference pattern for the counts per minute, or probability of arrival of electrons. Such an interference pattern would not be expected if the electrons behaved as classical particles, giving clear evidence that electrons are interfering, a distinct wave-like behavior. Figure 2.15: (a) Schematic of electron beam interference experiment, (b) Photograph of a double-slit interference pattern produced by electrons If we measure the angle θ at which the maximum intensity of the electrons arrives at the detector screen, we find they are described by exactly the same equation as that for light: 𝑑 𝑠𝑖𝑛 𝜃 = 𝑚 𝜆 , where m is the order number and λ is the electron wavelength. Therefore, the dual nature of the electron is clearly shown in this experiment: 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. UNCERTAINTY PRINCIPLE It is fundamentally impossible to make simultaneous measurements of a particle’s position and momentum with infinite accuracy. This is known as Heisenberg uncertainty principle. The uncertainties arise from the quantum structure of matter. 44 For a particle represented by a single wavelength wave existing throughout space,  is precisely known, and according to de Broglie hypothesis, its p is also known accurately. But the position of the particle in this case becomes completely uncertain. This means  = 0, p =0; but x =  In contrast, if a particle whose momentum is uncertain (combination of waves / a range of wavelengths are taken to form a wave packet), so that x is small, but  is large. If x is made zero,  and thereby p will become . In short ( x ) ( px) ≥ h / 4 (2.15) where x is uncertainty in the measurement of position x of the particle and px is uncertainty in the measurement of momentum px of the particle. One more relation expressing uncertainty principle is related to energy and time which is given by ( E ) ( t ) ≥ h / 4 (2.16) where E is uncertainty in the measurement of energy E of the system when the measurement is done over the time interval t. QUESTIONS 1. Explain (a) Stefan’s law (b) Wien’s displacement law (c) Rayleigh-Jeans law. 2. Sketch schematically the graph of wavelength vs intensity of radiation from a blackbody. 3. Explain Planck’s radiation law. 4. Write the assumptions made in Planck’s hypothesis of blackbody radiation. 5. Explain photoelectric effect. 6. What are the observations in the experiment on photoelectric effect? 7. What are the classical predictions about the photoelectric effect? 8. Explain Einstein’s photoelectric equation. 9. Which are the features of photoelectric effect-experiment explained by Einstein’s photoelectric equation? 10. Sketch schematically the following graphs with reference to the photoelectric effect: (a) photoelectric current vs applied voltage (b) kinetic energy of most-energetic electron vs frequency of incident light. 11. Explain Compton effect. 45 12. Explain the experiment on Compton effect. 13. Derive the Compton shift equation. 14. Explain the wave properties of the particles. 15. How Davison and Germer experiment validated the de Broglie hypothesis? Explain. 15. Explain a wave packet and represent it schematically. 16. Explain (a) group speed (b) phase speed, of a wave packet. 17. Show that the group speed of a wave packet is equal to the particle speed. 18. (a) Name any two phenomena which confirm the particle nature of light. (b) Name any two phenomena which confirm the wave nature of light. 19. Explain Heisenberg uncertainty principle. 20. Write the equations for uncertainty in (a) position and momentum (b) energy and time. 21. Mention two situations which can be well explained by the uncertainty relation. PROBLEMS 1 Find the peak wavelength of the blackbody radiation emitted by each of the following. A. The human body when the skin temperature is 35°C B. The tungsten filament of a light bulb, which operates at 2000 K C. The Sun, which has a surface temperature of about 5800 K. Ans: 9.4 μm, 1.4 μm, 0.50 μm 2 A 2.0- kg block is attached to a spring that has a force constant of k = 25 N/m. The spring is stretched 0.40 m from its equilibrium position and released. A. Find the total energy of the system and the frequency of oscillation according to classical calculations. B. Assuming that the energy is quantized, find the quantum number n for the system oscillating with this amplitude. C. Suppose the oscillator makes a transition from the n = 5.4 x 1033 state to the state corresponding to n = 5.4 x 1033 – 1. By how much does the energy of the oscillator change in this one-quantum change. Ans: 2.0 J, 0.56 Hz, 5.4 x 1033, 3.7 x 10–34 J 3 The human eye is most sensitive to 560 nm light. What is the temperature of a black body that would radiate most intensely at this wavelength? Ans: 5180 K 46  4 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 5 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 6 Calculate the energy in electron volts, of a photon whose frequency is (a) 620 THz, (b) 3.10 GHz, (c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons and state the classification of each on the electromagnetic spectrum. Ans: 2.57 eV, 1.28 x 10–5 eV, 1.91 x 10–7 eV, 484 nm, 9.68 cm, 6.52 m 7 An FM radio transmitter has a power output of 150 kW and operates at a frequency of 99.7 MHz. How many photons per second does the transmitter emit? Ans: 2.27 x 1030 photons/s 8 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. Ans: 1.67 eV, 504 nm 9 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? Ans: 296 nm, 1.01 x 1015 Hz, 2.71 V 10 Electrons are ejected from a metallic surface with speeds up to 4.60 x 105 m/s when light with a wavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cut-off frequency for this surface? Ans: 1.38 eV, 3.34 x 1014 Hz 11 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 47 12 Two light sources are used in a photoelectric experiment to determine the work function for a metal surface. When green light from a mercury lamp ( = 546.1 nm) is used, a stopping potential of 0.376 V reduces the photocurrent to zero. (a) Based on this what is the work function of this metal? (b) What stopping potential would be observed when using the yellow light from a helium discharge tube ( = 587.5 nm)? Ans: 1.90 eV, 0.215 V 13 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? Ans: 0.200710 nm, INCREASES 14 Calculate the energy and momentum of a photon of wavelength 700 nm. Ans: 1.78 eV, 9.47 x 10–28kg.m/s 15 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° 16 A 0.880 MeV photon is scattered by a free electron initially at rest such that the scattering angle of the scattered electron is equal to that of the scattered photon ( = ). (a) Determine the angles  & . (b) Determine the energy and momentum of the scattered electron and photon. Ans: 43°, 43°, 0.602 MeV, 3.21 x 10–22 kg.m/s, 0.278 MeV, 3.21 x 10–22 kg.m/s 17 Calculate the de- Broglie wavelength for an electron moving at 1.0 x 107 m/s. Ans: 7.28 x 10–11 m 18 A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength? Ans: 3.3 x 10–34 m 19 A particle of charge q and mass m has been accelerated from rest to a nonrelativistic speed through a potential difference of V. Find an expression for its de Broglie wavelength. h Ans: λ = √2 m q Δv 20 (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 21 In the Davisson-Germer experiment, 54.0 eV electrons were diffracted from a nickel lattice. If the first maximum in the diffraction pattern was observed at = 50.0°, what was the lattice spacing a between the vertical rows of atoms in the figure? 48  Ans: 2.18 x 10–10 m 22 Consider a freely moving quantum particle with mass m and speed u. Its energy is E= K= mu2/2. Determine the phase speed of the quantum wave representing the particle and show that it is different from the speed at which the particle transports mass and energy. Ans: vGROUP = u ≠ vPHASE 23 Electrons are incident on a pair of narrow slits 0.060 m apart. The ‘bright bands’ in the interference pattern are separated by 0.40 mm on a ‘screen’ 20.0 cm from the slits. Determine the potential difference through which the electrons were accelerated to give this pattern. Ans: 105 V 24 The speed of an electron is measured to be 5.00 x 103 m/s to an accuracy of 0.0030%. Find the minimum uncertainty in determining the position of this electron. Ans: 0.383 mm 25 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 26 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 27 Find the minimum kinetic energy of a proton confined within a nucleus having a diameter of 1.0 x 10–15 m. Ans: 5.2 MeV 49  3 QUANTUM MECHANICS OBJECTIVES: To learn the application of Schrödinger equation to a bound particle and to learn the quantized nature of the bound particle, its expectation values and physical significance. To understand the tunnelling behaviour of a particle incident on a potential barrier. To understand the behaviour of quantum oscillator. AN INTERPRETATION OF QUANTUM MECHANICS Experimental evidences proved that both matter and electromagnetic radiation exhibit wave and particle nature depending on the phenomenon being observed. Making a conceptual connection between particles and waves, for an electromagnetic radiation of amplitude E, the probability per unit volume of finding a photon in a given region of space at an instant of time as; PROBABILITY ∝ 𝐸2 𝑉 Figure 3.1: Wave packet Taking the analogy between electromagnetic radiation and matter-the probability per unit volume of finding the particle is proportional to the square of the amplitude of a wave representing the particle, even if the amplitude of the de Broglie wave associated with a particle is generally not a measurable quantity. The amplitude of the de Broglie wave associated with a particle is called probability amplitude, or the wave function, and is denoted by . In general, the complete wave function  for a system depends on the positions of all the particles in the system and on time. This can be written as (r1,r2,…rj,…,t) = (rj) e–it where rj is the position vector of the jth particle in the system. 50 For any system in which the potential energy is time-independent and depends only on the position of particles within the system, the important information about the system is contained within the space part of the wave function. The wave function  contains within it all the information that can be known about the particle. | |2 is always real and positive, and is proportional to the probability per unit volume, of finding the particle at a given point at some instant. If  represents a single particle, then ||2 - called the probability density - is the relative probability per unit volume that the particle will be found at any given point in the volume. One-dimensional wave functions and expectation values: Let  be the wave function for a particle moving along the x axis. Then P(x) dx = ||2dx is the probability to find the particle in the infinitesimal interval dx around the point x. The probability of finding the particle in the arbitrary interval a ≤ x ≤ b is 𝑏 𝑃𝑎𝑏 = ∫𝑎 |𝜓|2 𝑑𝑥 (3.1) The probability of a particle being in the interval a ≤ x ≤ b is the area under the probability density curve from a to b. The total probability of finding the particle is one. Forcing this condition on the wave function is called normalization. +∞ ∫−∞ |𝜓|2 𝑑𝑥 = 1 (3.2) Figure 3.2: An arbitrary probability density curve for a particle All the measurable quantities of a particle, such as its position, momentum and energy can be derived from the knowledge of . e.g., the average position at which one expects to find the particle after many measurements is called the expectation value of x and is defined by the equation +∞ 〈𝑥〉 ≡ ∫−∞ 𝜓 ∗ 𝑥 𝜓 𝑑𝑥 (3.3) 51 The important mathematical features of a physically reasonable wave function (x) for a system are (x) may be a complex function or a real function, depending on the system. (x) must be finite, continuous and single valued everywhere. The space derivatives of, must be finite, continuous and single valued everywhere.  must be normalizable. THE SCHRÖDINGER EQUATION The appropriate wave equation for matter waves was developed by Schrödinger. Schrödinger equation as it applies to a particle of mass m confined to move along x axis and interacting with its environment through a potential energy function U(x) is ℏ2 𝑑2 𝜓 − +𝑈𝜓 = 𝐸𝜓 (3.4) 2 𝑚 𝑑𝑥 2 where E is a constant equal to the total energy of the system (the particle and its environment) and ħ = h/2. This equation is referred to as the one dimensional, time-independent Schrödinger equation. Application of Schrödinger equation: 1. Particle in an infinite potential well (particle in a box) 2. Particle in a finite potential well 3. Tunnelling 4. Quantum oscillator PARTICLE IN AN INFINITE POTENTIAL WELL (PARTICLE IN A “BOX”) Figure 3.3: (a) Particle in a potential well of infinite height, (b) Sketch of potential well 52 Consider a particle of mass m and velocity v, confined to bounce between two impenetrable walls separated by a distance L, as shown in Figure 3.3(a). Figure 3.3(b) shows the potential energy function for the system. U(x) = 0, for 0