www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS UNIT-1 MODERN PHYSICS Introduction to blackbody radiation spectrum: A body which absorbs all radiation that is incident on it is called a perfect blackbody. When radiation allowed to fall on such a body, it is neither reflected nor transmitted. Such a black body after absorbing the incident radiation gets heated and starts emitting radiation of all possible wavelengths. In practice, perfect blackbody does not exist and we can have objects that are only close to a blackbody. A blackbody, on heating, can emit all radiations it has absorbed and is called blackbody radiation. Figure shows the energy distribution curves in which energy density E (energy emitted by the blackbody per unit area of the surface) is plotted as a function of wavelength at different temperatures of the blackbody. The important points that can be noted down from these curves are 1. All curves shows a peak suggesting that the emitted intensity is maximum at a particular wavelength. 2. An increase in temperature results in an increase in the energy emitted. 3 .As the temperature increases, the peak shifts to lower wavelength. Laws of blackbody radiation: 1. Stefan-Boltzmann law It states that the total energy density E0 of radiation emitted from a blackbody is directly proportional to the fourth power of its absolute temperature T. i.e., E0 T4 Or E0 = T4 where is a constant called Stefan’s constant with numerical value equal to 5.67×10-8 Wm-2K-4. This law agrees well with the experimental results. 2. Wien’s displacement law It states that the wavelength m corresponding to the maximum emissive energy decreases with increasing temperature. i.e., m T Or m T = b, where b is called the Wien’s constant and is equal to 2.9×10-3 mK. 3. Wien’s law Using laws of thermodynamics and classical concepts, Wien developed an expression for energy density Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 1 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS as, () = where C1 and C2 are constants. This law holds good for smaller values of wavelength. 4. Rayleigh-Jeans law Rayleigh derived an expression for the energy density of radiation based on classical theory which is given by, () = where k is called Boltzmann’s constant and its value is 1.381 x 10-23JK-1 This law holds good only for large values of wavelength. As per the Rayleigh-Jeans law, the radiant energy increases with decreasing wavelength and a blackbody must radiate all the energy at very short wavelength. But in actual practice, it doesn’t happen so. The failure of Rayleigh-Jeans law to explain the aspect of very little emission of radiation beyond the violet region towards the lower wavelength side of the spectrum is referred as ultraviolet catastrophe. Both Wien’s law and Rayleigh-Jeans law indicates failure of classical theory in explaining blackbody radiation. Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 2 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Planck’s radiation law: Max Planck proposed a law based on quantum theory. According to this, atoms or molecules absorb or emit radiation in quanta or small energy packets called photons. If ‘’ is the frequency of photons, then its energy can be explained as E=h where h=6.63x10-34Js is called Planck’s constant. Applying quantum theory, Planck obtained an expression for energy density of blackbody radiation as, () = − ----------------- (1) This law agrees well with the experimental observation of blackbody radiation and is valid for all wavelengths. 1. For shorter wavelengths: i.e., when λ is small, is very large Or i.e., >>1 hence ( − ) ≈ Substituting this in Planck’s radiation law i.e., in eq(1) then, () = i.e., () = hc Where C1=8hc and C2= k Hence at smaller wavelengths, Planck’s radiation law reduces to Wien’s law. 2. For longer wavelengths: i.e., when λ is large, is very small Expanding the power series , we have Since =+ ! + ) ( ! +−−−−−−− is very small, its higher power terms can be neglected Then the above expression becomes = + hc λkT or – = hc λkT Substituting this in Planck’s radiation law i.e., in eq(1), () = i.e., () = Hence at longer wavelengths, Planck’s radiation law reduces to Rayleigh-Jeans law. Photoelectric effect: Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 3 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Emission of electrons from a metal surface when light of suitable energy falls on it is called photoelectric effect. The experimental setup for observing photoelectric effect consists of a pair of metal plate electrodes in an evacuated discharge tube connected to a voltage source as shown: When light of suitable energy is incident on the cathode, electrons are emitted and a current flows across the discharge tube. Some special features of photoelectric emission are: 1. It is an instantaneous process-there is no time interval between the incidence of light and the emission of photoelectrons. 2. There is a minimum frequency for the incident light called threshold frequency, below which no photoelectric emission occurs. This depends upon the nature of the material of the emitter surface. The energy corresponding to the threshold frequency, called the work function is the minimum energy required to release an electron from the emitter surface. 3. For a given frequency of the incident light, photocurrent is directly proportional to the intensity of the incident light. 4. The photoelectron emission can be stopped by applying a reverse voltage to the phototube. ie., by making the emitter electrode positive and the collector negative. The negative collector potential required to stop the photoelectron emission is called the stopping potential. Einstein’s Theory: Photoelectric effect can be explained on the basis of quantum theory of light. When the energy equal to work function of the metal is incident on the metal surface, the incident photon liberates electrons from their bound state. When the incident photon carries energy in excess of the work function, the extra energy appears as the kinetic energy of the emitted electron. When the intensity of light increases, the number of photoelectrons emitted increases but their kinetic energy remain unaltered. When a photon of frequency ‘’ is incident on a metal surface of work function ‘’, then, h = + ( mv2)max 1 where ( 2 mv2)max is the kinetic energy of the emitted photoelectrons. This is known as Einstein’s photoelectric equation. Since = h0, it can also be written as, Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 4 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS 1 ( 2 mv2)max = h - = h(-0) If V0 is the stopping potential corresponding to the incident photon frequency , then, 1 ( 2 mv2)max = h - 1 ( 2 mv2)max = h(-0) or ( mv2)max = eV0 Compton Effect: When X-rays are scattered by a solid medium, in addition to the scattered X-rays of same frequency, there exist some scattered X-rays of a slightly lower frequency (higher wavelength). Compton observed this phenomenon and is called Compton Effect. Compton Effect can be explained on the basis of the quantum theory and laws of conservation of energy and momentum. Consider an x-ray photon of energy h incident on an electron at rest. After the interaction, the X-ray photon gets scattered at an angle with its energy changed to h’ and the electron which was initially at rest recoils at an angle . It can be shown that the increase in wavelength is given by h ∆λ= (1- cos θ) m0 C h When = 900, = m 0 where m0 is the rest mass of the electron. = 0.0242Å. This constant value is called Compton wavelength. C Wave particle dualism: The Photoelectric Effect and Compton Effect conclusively established the particle behavior of light. The phenomena of interference, diffraction and polarization give exclusive evidence for the wave behavior of light. Hence we have to conclude that light behaves as an advancing wave in some phenomena and it behaves as a flux of particles in some other phenomena. Therefore we say that light exhibits wave-particle duality. De-Broglie’s hypothesis: De-Broglie extended the wave particle dualism of light to the material particles. This is known as deBroglie hypothesis. According to this hypothesis, material particles in motion possess a wave character. The waves associated with material particles are called matter waves or de-Broglie waves. According to Planck’s theory of radiation, E = h --------- (1) where is the frequency associated with the radiation. According to Einstein’s mass-energy relation, E = mc2 --------- (2) where m is the mass of the photon and c is the velocity of light Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 5 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Combining (1) and (2), hc i.e., h = mc2 => = mc 2 (since = ) λ h = mc λ Therefore momentum associated with the particle is given by p = mc, Or, = where is called de-Broglie wavelength. De-Broglie wavelength associated with the accelerated electron: A beam of high energy electrons can be obtained by accelerating them in an electric field. Consider an electron starting from rest when accelerated with a potential difference V, the kinetic energy (E) acquired by the electron is given by, 1 E= mv2 and also E= eV 2 Thus, Or 1 mv 2 2 m2 v 2 2m = eV = eV p2 i.e., =eV =E 2m where ‘v’ is the velocity of the electron, ‘m’ its mass and ‘p’ the momentum. Now the momentum may be expressed as, p = √2mE = √2meV Hence the de-Broglie wavelength λ= h p = h √2mE = h √2meV To test de-Broglie hypothesis, Heisenberg and Schrödinger formulated theories whereas G.P.Thomson, Davisson and Germer conducted experiments. Davisson-Germer experiment: The electron diffraction experimental setup used by Davisson and Germer to verify de-Broglie’s hypothesis is as shown: The filament F is heated to produce electrons via thermionic emission. These electrons are passed through a narrow aperture forming a fine beam of accelerated electrons. The electron beam was then made to incident on a single crystalline sample of Nickel. The electrons scattered at different angles were counted using a detector. The experiment was repeated by recording the scattered electron intensities at various positions of the detector. Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 6 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS A sharp maximum occurred in electron density at an angle (Φ) of 500 with the incident beam for an accelerating potential of 54 V. The angle of incidence corresponding to this is 250 and from figure, glancing angle θ (angle of diffraction) is 650. From X-ray diffraction experiment, spacing of the planes responsible for diffraction was found to be 0.091 nm. Assuming first order diffraction, Bragg’s law can be written as, = 2d sin = 2×0.091×10-9×sin65 = 0.165 nm. By de-Broglie relation, = √ = . × − √(×.×− ×.×− ×) = 0.167nm Thus Davisson and Germer experiment directly verifies the de-Broglie hypothesis. Characteristic properties of matter waves: 1. Matter waves are associated with moving particle. 2. Wavelength of matter waves is inversely proportional to the velocity with which the particle is moving h ( =mv). Hence a particle at rest has an infinite wavelength. 3. Wavelength of matter waves is inversely proportional to the mass of the particle. Hence wavelike behavior of heavier bodies is not very evident whereas wave nature of subatomic particles could be observed experimentally. 4. Wave function is used to define a matter wave which is related to the probability of finding a particle at any place at any instant. 5. Matter waves are represented by a wave packet made up of a group of waves of slightly differing wavelengths. Hence we talk of group velocity of matter waves rather than the phase velocity (velocity of a single wave). The group velocity can be shown to be equal to the particle velocity. Phase velocity: General expression for a wave is Y = A cos (t-kx) where Y = Displacement at any instant t, A = Amplitude of vibration, = 2 is the angular frequency and 2 k = is the wave vector or wave number. Phase velocity or wave velocity of a wave is the velocity of the wave when phase is constant. i.e., t-kx = constant or, kx = t + constant ωt or, x = k + constant Hence Phase velocity vp = = Group velocity: The de-Broglie waves are represented by a wave packet and hence we have ‘group velocity’ associated with them. Group velocity is the velocity with which the wave packet travels. Consider two waves having same amplitude but having slightly different frequency and wave number represented by the equations Y1 = A cos (ωt-kx) Y2 = A cos [(ω+ Δω) t – (k+ Δk) x] The resultant displacement due to the superposition of the above two waves is, Y = Y1 + Y2 = A cos (ωt-kx) + A cos [(ω+ Δω) t – (k+ Δk)x] Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 7 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS A B A B Since, cos A + cos B = 2cos ( ) cos ( ), 2 2 2 2k k k Y = 2A cos {( )t–( ) x} cos {( )t ( ) x} 2 2 2 2 As the difference in frequency of the two waves is very small, we can assume, 2+Δω 2 and 2k+Δk 2k k Y = 2Acos {( )t ( ) x} cos (ωt-kx) 2 2 The velocity of the resultant wave (group velocity) is given by the speed with which a reference point, say the maximum amplitude point, moves. Taking the amplitude of the resultant wave as constant, we have, k 2Acos {( )t ( ) x} = constant 2 2 k or, ( )t ( ) x = constant 2 2 or, x = ∆ ∆ + constant ∆ Group velocity vg = = ∆ When and k are very small, vg = Relation between group velocity and phase velocity: Phase velocity, vp = where ω is the angular frequency of the wave = vp k ----- (1) Group velocity, vg = vg = dvp k (From eq (1)) dk dk dvp 2 dk dvp = vp dk + k = vp + ( ) 1 1 d( ) dvp dλ 1 1 d( ) dλ dvp dλ = vp + ( ) = vp + ( ) = vp + ( ) But 1 λ d( ) dλ = where the propagation constant (or wave number) k = 2 d( ) dvp 2 1 dλ d(1) −1 λ2 1 dvp Therefore vg= vp+ () dλ λ2 ( −1) Thus vg= vp- Relation between group velocity and particle velocity: E Energy of a photon E = hν or ν = h ------ (1) Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 8 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS (2E) We know the angular frequency of the wave ω = 2ν or ω = h 2 dω = ( h ) dE ------(2) 2 Further, k = = λ 2p since = h h p 2 dk = ( h ) dp ------(3) Substituting the value of d and dk from equations (2) and (3) in the expression for group velocity, dω dE vg = dk = dp ------ (4) If a particle of mass m is moving with a velocity vparticle, its energy is given by, 2 1 E = 2 mv2particle = 2 -------(5) Substituting this in equation (4), dE vg = dp = 2 2 d p dp =m mVparticle = = vparticle m Hence vg = vparticle Relation between velocity of light, group velocity and phase velocity: Phase velocity, vp = where k is the propagation constant or wave number We know the angular frequency of the wave ω = 2ν or ω = Further, Wave number k = Thus vp = = 2 = λ 2p since = h h p (2E) h where p is the momentum of the wave (2E) h 2p h E =p mc2 (Since E = mc2) = mv =v particle c2 particle vphase × vparticle = c2 But vg = vparticle vphase × vg = c2 Expression for de-Broglie wavelength using group velocity: We know, the group velocity vg = dω , where angular frequency ω = 2 and wave number k = dk 1 2 λ dω = 2d and dk = 2 d( λ ) Thus vg = dω = dk 1 2d 1 2 d( ) λ Or, d( λ ) = d vg = = d 1 d( ) λ d vparticleg -------- (1) [since vg = vparticle] Total energy of a particle moving under an applied potential V is given by, 1 E = 2 mv2 + V Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 9 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS de-Broglie related E = h to above expression 1 Then, h = mv2 + V 2 Assuming V as a constant potential and differentiating the above equation, h d = m vparticle dv mv or, d = ( particle ) dv h Substituting this in equation (1), 1 m dv d( λ) = h Integrating, 1 mv = h + constant λ i.e., 1 λ = p h + constant Assuming constant of integration to be zero, 1 λ = or, λ = p h + constant , the de-Broglie wavelength. ************** VTU Model Question Paper 1 .a) 1) If the momentum of a particle is increased to four times, the de-Broglie wavelength is i) become twice ii) become for times iii) become one-fourth iv) become half 2) Blackbody radiation spectrum, maximum intensity is shifting towards i) shorter wavelength ii) longer wavelength iii) no change iv) none of these 3) Group velocity of wave is equal to i) V phase ii) V particle iii) Velocity of light iv) none of these 4) de-Broglie wavelength of an electron accelerated by a potential of 60 V is i) 1.85 Å ii) 1.58 Å iii) 1.589 Å iv) 1.57 Å (4 marks) b) Describe Davisson-Germer experiment to prove the dual nature of matter waves. (8marks) c) Explain phase velocity and Group velocity. Derive de-Broglie wavelength using Group velocity. (8 marks) Dec 08/ Jan 09 1 a) 1) The de-Broglie wavelength associated with an electron of mass m and accelerated by a potential V is h √2mVe h h i) ii) ℎ iii) Vem iv) 2Vem √2mVe 2) Davisson and Germer were the first to demonstrate: i) The straight line propagation of light ii) The diffraction of photons iii) The effective mass of electron iv) None of these 3) Electrons behaves as waves because they can be: i) Deflected by an electric field ii) Diffracted by a crystal iii) Deflected by magnetic field iv) They ionize a gas 4) In Davisson-Germer experiment, the hump is most prominent when the electron is accelerated by i) 34 volts ii) 54 volts iii) 60 volts iv) 80 volts (04 Marks) b) Define phase velocity and group velocity. Show that group velocity is same as particle velocity. (08 Marks) c) Derive de-Broglie wavelength using Group velocity. (04 Marks) d) Compare the energy of a photon with that of a neutron when both are associated with wavelength of 1 Å. Given that mass of neutron is 1.678 × 10 -27 kg. (04 Marks) June-July 2009 1) a) 1) An electron and a proton are accelerated through same potential. The ratio of de-Broglie wavelength Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 10 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS e/p is i) 1 ii) me iii) mp mp mp iv) √ me me 2) Wavefunction associated with a material particle is i) Single valued ii) Finite iii) Continuous iv) all the above 3) In a blackbody radiation spectrum, the maximum energy peaks shift towards the shorter wavelength side with the increase in temperature. This confirms i) Stefan’s law ii) Wein’s law iii) Rayleigh-Jean’s law iv) Planck’s law 6 4) The group velocity of the particle is 3×10 m/s, whose phase velocity is i) 6.06×106 m/s ii) 3×1010 m/s iii) 3 nm/s iv) 1.5×1010 m/s (04 Marks) b) Describe Davisson and Germer experiment for confirmation of deBroglie hypothesis. (08 Marks) c) Explain phase and group velocity. Calculate the de-Broglie wavelength of a bullet of mass 5 gm moving with a velocity 20 km/hr. (08 Marks) Dec.09/Jan.10 1 a)1) Wien’s law is deduced from Planck’s radiation formula under the condition of i) Very small wavelength and temperature ii) Large wavelength and temperature iii) Small wavelength and high temperature iv) Large wavelength and small temperature 2) The Compton wavelength is given by i) h/m0C2 ii) h2/m0C2 iii) h/m0C iv) h2/2m0C 3) Which of the following relations can be used to determine de-Broglie wavelength associated with a particle? h h h i) ii) Vm iii) iv) All of these √2mE √2mVe 6 4) If the group velocity of a particle is 3×10 m/s, its phase velocity is i) 100 m/s ii) 3×106 m/s iii) 3×108 m/s iv) 3×1010 m/s (04 Marks) b) What is Planck’s radiation law? Show how Wien’s law and Rayleigh-Jean’s law can be derived from it. (06 Marks) c) Define group velocity. Derive relation between group velocity and phase velocity. (06 Marks) d) A fast moving neutron is found to be have a associated de-Broglie wavelength 2Å. Find its kinetic energy and group velocity of the de-Broglie waves. (04 Marks) May/June 2010 1 a) 1) In a blackbody radiation spectrum, the Wien’s distribution law is applicable only for i) Longer wavelength ii) Shorter wavelength iii) Entire wavelength iv) None of these 2) The de-Broglie wavelength associated with an electron of mass m and accelerated by a potential V is h √2mVe h h i) ii) ℎ iii) Vem iv) 2Vem √2mVe 3) Electrons behaves as a wave because they can be i) Diffracted by a crystal ii) Deflected by magnetic field iii) Deflected by electric field iv) Ionise a gas 4) If the group velocity of de-Broglie wave is 4×108 m/sec, its phase velocity is i) 12×108 m/sec ii) 2.25×108 m/sec iii) 5.33×108 m/sec iv) 1.33×108 m/sec (04 Marks) b) Explain duality of matter waves. (04 Marks) c) Define phase velocity and group velocity. Show that group velocity is equal to particle velocity. (08 Marks) d) Calculate the momentum of the particle and de-Broglie wavelength associated with an electron with a kinetic energy of 1.5 keV. (04 Marks) January 2011 Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 11 www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS 1 a) 1) Green light incident on a surface releases photoelectrons from the surface. If now blue light is incident on the same surface, the velocity of electrons i) increases ii) decreases iii) remains same iv) becomes zero 2) Rayleigh-Jean’s theory of radiations agree with experimental results for i) all wavelengths ii) shorter wavelengths only iii) longer wavelengths only iv) middle order wavelengths only 3) The de-Broglie wavelength of an electron accelerated to a potential difference of 100 volts is i) 1.2Å ii) 10Å iii) 100Å iv) 12Å 4) The wave nature associated with electrons in motion was verified by i) photoelectric effect ii) Compton effect iii) diffraction by crystals iv) Raman effect (04Marks) b) State and explain de-Broglie’s hypothesis. (04 Marks) c) Define phase velocity and group velocity. Obtain the relation between group velocity and particle velocity. Obtain the expression for de-Broglie wavelength using group velocity. (08 Marks) d) Find the kinetic energy and group velocity of an electron with de-Broglie wavelength of 0.2 nm. (04 marks) ************* Notes Compiled by: Dr. Santhosh D Shenoy, M.Sc., Ph.D. www.bookspar.com | Website for Students | VTU NOTES | QUESTION PAPERS | NEWS | RESULTS Page 12

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