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# Quantum Mechanics And Statistical Mechanics PHYSICS 1251

University of Petroleum and Energy Studies

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Quantum Mechanics Introduction There are a few phenomenon which the classical mechanics failed to explain. 1. Stability of an atom 2. Spectral series of Hydrogen atom 3. Black body radiation Max Planck in 1900 at a meeting of German Physical Society read his paper “On the theory of the Energy distribution law of the Normal Spectrum”. This was the start of the revolution of Physics i.e. the start of Quantum Mechanics. Quantum Mechanics It is a generalization of Classical Physics that includes classical laws as special cases. Quantum Physics extends that range to the region of small dimensions. Just as ‘c’ the velocity of light signifies universal constant, the Planck's constant characterizes Quantum Physics. h 6.6510 27erg.sec h 6.62510 34Joule.sec Quantum Mechanics It is able to explain 1. Photo electric effect 2. Black body radiation 3. Compton effect 4. Emission of line spectra The most outstanding development in modern science was the conception of Quantum Mechanics in 1925. This new approach was highly successful in explaining about the behavior of atoms, molecules and nuclei. Photo Electric Effect The emission of electrons from a metal plate when illuminated by light or any other radiation of any wavelength or frequency (suitable) is called photoelectric effect. The emitted electrons are called ‘photo electrons’. Light Evacuated Quartz tube Metal Collecting _ + plate plate V ^^^^^^^^ A Photo Electric Effect Experimental findings of the photoelectric effect 1. There is no time lag between the arrival of light at the metal surface and the emission of photoelectrons. 2. When the voltage is increased to a certain value say V , the o photocurrent reduces to zero. 3. Increase in intensity increase the number of the photoelectrons but the electron energy remains the same. 3I Photo 2I Current I Vo Voltage Photo Electric Effect 4. Increase in frequency of light increases the energy of the electrons. At frequencies below a certain critical frequency (characteristics of each particular metal), no electron is emitted. Photo Current v1 v2v 3 Voltage Einstein’s Photo Electric Explanation The energy of a incident photon is utilized in two ways 1. A part of energy is used to free the electron from the atom known as photoelectric workfunction (o ). 2. Other part is used in providing kinetic energy to the emitted electron . 1mv2 2 1 h W mv 2 o 2 This is called Einstein’s photoelectric equation. h W KE o max h h o KE max KE max h( o) If o, no photoelectric effect hc W o h o o hc 12400 o o A W o W oV ) IfVo is the stopping potential, then KE max h( o eVo h h o h h o Vo e e It is in formy mxc . The graph witho on y-axis and on x-axis will be a straight line wh e slope Photons to explain detailed results of photoelectric experiment. photon, hc E p h Photon has zero rest mass, travels at speed of light Explains “instantaneous” emission of electrons in photoelectric effect, frequency dependence. Compton Effect When a monochromatic beam of X-rays is scattered from a material then both the wavelength of primary radiation (unmodified radiation) and the radiation of higher wavelength (modified radiation) are found to be present in the scattered radiation. Presence of modified radiation in scattered X-rays is called Compton effect. E' h' h ' sin scattered incident c photon E h photon mvcos h ' h electron c cos p v c mvsin recoiled electron From Theory of Relativity, total energy of the recoiled electron with v ~ c is E mc K m c 2 o 2 2 K mc m c o m c 2 K o m o 2 1 v c2 2 2 1 K m o 2 2 1 1v c Similarly, momentum of recoiled electron is m o mv 2 2 1 v c Now from Energy Conservation h h m c 2 1 1 (i) o 1v c 2 From Momentum Conservation h h' m o cos 2 2cos (ii) along x-axis c c 1v c and 0 h'sin mov sin (iii) along y-axis c 1 v c 2 Rearranging (ii) and squaring both sides 2 h h ' m o 2 2 cos 2 2cos (iv) c c 1 v c Rearranging (iii) and squaring both sides 2 2 2 h ' m o 2 sin 2 2sin (v) c 1 c Adding (iv) and (v) 2 2 2 2 2 h h' 2h ' mov c c c 2 cos 1v c 2 (vi) From equation (i) h h' moc m o 2 2 c c 1v c On squaring, we get h 2 h' 2 2 2 2h2 ' m c 2 m co 2 2hm (o o2 2 c c c 1v c (vii) Subtracting (vi) from (vii) 2 2 h '(1cos)2 hm ( ') 0 c2 o 2 2hm ( ') 2h '(1cos) o c 2 h ' m o ') 2 (1cos) c c c But and So, ' 1 1 h moc (1cos) ' ' m c h (1cos) o ' ' h ' m c (1cos) o is the Compton Shift. It neither depends on the incident wavelength nor on the scattering material. It only on the scattering ange i.e. h is called the Compton wavelength of the electron m o and its value is 0.0243 Å. Experimental Verification Monochromatic photon Bragg’s X-ray X-ray Source Spectrometer θ Graphite target 1. One peak is found at same position. This is unmodified radiation 2. Other peak is found at higher wavelength. This is modified signal of low energy. 3. increases with increase i. Compton effect can’t observed in Visible Light h (1cos) 0.0243 (1- cosθ) Å moc is maximum when (1- cosθ) is maximum i.e. 2. max 0.05 Å So Compton effect can be observed only for radiation having wavelength of few Å. For 1Å ~ 1% For 5000Å ~ 0.001% (undetectable) Pair Production When a photon (electromagnetic energy) of sufficient energy passes near the field of nucleus, it materializes into an electron and positron. This phenomenon is known as pair production. e Photon Nucleus (+ve) e In this process charge, energy and momentum remains conserved prior and after the production of pair. The rest mass energy of an electron or positron is 0.51 MeV (according to E = mc ). The minimum energy required for pair production is 1.02 MeV. Any additional photon energy becomes the kinetic energy of the electron and positron. The corresponding maximum photon wavelength is 1.2 pm. Electromagnetic waves with such wavelengths are called gamma rays () . Pair Annihilation When an electron and positron interact with each other due to their opposite charge, both the particle can annihilate converting their mass into electromagnetic energy in the form of two - rays photon. e e Charge, energy and momentum are again conversed. Two - photons are produced (each of energy 0.51 MeV plus half the K.E. of the particles) to conserve the momentum. Pair production cannot occur in empty space From conservation of energy h 2m c 2 o 2 2 here mois the rest mass and 1 1v c e p pcos h c pcos p e In the direction of motion of the photon, the momentum is conserved if h 2pcos c h 2cpcos (i) Momentum of electron and positron is p mov Equation (i) now becomes h 2m ov cos 2 h m c o cos But vc 1 and cos 1 h 2m co 2 But conservation of energy requires that h 2m co Hence it is impossible for pair production to conserve both the energy and momentum unless some other object is involved in the process to carry away part of the initial photon momentum. Therefore pair production cannot occur in empty space. Wave Particle Duality Light can exhibit both kind of nature of waves and particles so the light shows wave-particle dual nature. In some cases like interference, diffraction and polarization it behaves as wave while in other cases like photoelectric and compton effect it behaves as particles (photon). De Broglie Waves Not only the light but every materialistic particle such as electron, proton or even the heavier object exhibits wave- particle dual nature. De-Broglie proposed that a moving particle, whatever its nature, has waves associated with it. These waves are called “matter waves”. Energy of a photon is E h For a particle, say photon of mass, m E mc 2 2 mc hv 2 hc mc h mc Suppose a particle of mass, m is moving with velocity, v then the wavelength associated with it can be given by h h or p mv (i) If 0 means that waves are associated with moving material particles only. (ii) De-Broglie wave does not depend on whether the moving particle is charged or uncharged. It means matter waves are not electromagnetic in nature. Wave Velocity or Phase Velocity When a monochromatic wave travels through a medium, its velocity of advancement in the medium is called the wave velocity or phase velocipy (V ). V p k where 2 is the angular frequency and k 2 is the wave number. Group Velocity In practice, we came across pulses rather than monochromatic waves. A pulse consists of a number of waves differing slightly from one another in frequency. The observed velocity is, however, the velocity with which the maximum amplitude of the group advances in a medium. So, the group velocity is the velocity with which the energy in the group is transmittedg(V ). The individual waves travel “inside” the group with their phase velocities. d Vg dk Relation between Phase and Group Velocity d d Vg (kVp) dk dk V V k dVp g p dk 2 dV p V g Vp d 2 1 dV p Vg Vp d 1 1 dVp Vg V p 1 2d dVp Vg V p d In a Dispersive medium V depends on frequency p i.e. constant k dV So, p is positive generally (not always). d V g p generally dVp Vg Vp d In a non-dispersive medium ( such as empty space) constant V p k dVp 0 d V V g p Phase Velocity of De-Broglie’s waves According to De-Broglie’s hypothesis of matter waves h mv 2 2mv wave number k h (i) If a particle has energy E, then corresponding wave will have frequency E h 2 2E then angular frequency will be h 2 2 mc (ii) h Dividing (ii) by () 2 mc 2 h k h 2mv 2 V c p v But v is always < c (velocity of light) (i) Velocity of De-Broglie’s wavep c (not acceptable) (ii) De-Broglie’s waves (Vp) will move faster than the particle velocity (v) and hence the waves would left the particle behind. Group Velocity of De-Broglie’s waves The discrepancy is resolved by postulating that a moving particle is associated with a “wave packet” or “wave group”, rather than a single wave-train. A wave group having wavelength λ is composed of a number of component waves with slightly different wavelengths in the neighborhood of λ. Suppose a particle of rest massom moving with velocity v then associated matter wave will have 2 2 mc 2mv m m o and k where 1v c 2 h h 2m o 2 2 m v 2 2 and k o h 1 v c h 1v c2 2 On differentiating w.r.t. velocity, v d 2m o dv 2 2 2 (i) h1 c dk 2m o 2 2 2 (ii) dv h1 c Dividing (i) by (ii) d dv 2m o dv dk 2m o d v V dk g Wave group associated with a moving particle also moves with the velocity of the particle. Moving particl wave packet or wave group Davisson & Germer experiment of electron diffraction • If particles have a wave nature, then under appropriate conditions, they should exhibit diffraction • Davisson & Germer measured the wavelength of electrons • This provided experimental confirmation of the matter waves proposed by de Broglie Davisson and Germer Experiment 0 90 Incident Beam 0 Current vs accelerating voltage has a maximum (a bump or kink noticed in the graph), i.e. the highest number of electrons is scattered in a specific direction. The bump becomes most prominent for 54 V at φ ~ 50° According to de Broglie, the wavelength associated with an electron accelerated through V volts is o 12.28 A V Hence the wavelength for 54 V electron o 12.28 1.67 A 54 From X-ray analysis we know that the nickel crystal acts as a plane diffraction grating with grating space d = 0.91 Å Here the diffraction angle, φ ~ 50° The angle of incidence relative to the family of Bragg’s plane 180o 50 o o 65 2 From the Bragg’s equation 2d sin o o 2(0.91A)sin65 1.65 A which is equivalent to the λ calculated by de-Broglie’s hypothesis. It confirms the wavelike nature of electrons Electron Microscope: Instrumental Application of Matter Waves Resolving power of any optical instrument is proportional to the wavelength of whatever (radiation or particle) is used to illuminate the sample. An optical microscope uses visible light and gives 500x magnification/200 nm resolution. Fast electron in electron microscope, however, have much shorter wavelength than those of visible light and hence a resolution of ~0.1 nm/magnification 1,000,000x can be achieved in an Electron Microscope. Heisenberg Uncertainty Principle It states that only one of the “position” or “momentum” can be measured accurately at a single moment within the instrumental limit. or It is impossible to measure both the position and momentum simultaneously with unlimited accuracy. x uncertainty in position p x uncertainty in momentum then h x p x 2 2 The product ofx &p xof an object is greater than or equal to 2 If x is measured accuratelyx 0 p x The principle applies to all canonically conjugate pairs of quantities in which measurement of one quantity affects the capacity to measure the other. Like, energy E and time t. E t 2 and angular momentum L and angular position θ 2 Applications of Heisenberg Uncertainty Principle (i) Non-existence of electron in nucleus Order of radius of an nucleus ~ 5mx10 If electron exist in the nucleus then 510 1m max x p x 2 (x)max(px min 2 20 1 (px min 2x 1.110 kg. . then E pc 20MeV relativistic Thus the kinetic energy of an electron must be greater than 20 MeV to be a part of nucleus Experiments show that the electrons emitted by certain unstable nuclei don’t have energy greater than 3-4 MeV. Thus we can conclude that the electrons cannot be present within nuclei. Concept of Bohr Orbit violates Uncertainty Principle x.p 2 p2 But E 2m pp mvp x E p m m t E.t x.p E.t 2 According to the concept of Bohr orbit, energy of an electron in a orbit is constant i.e. ΔE = 0. E.t 2 t All energy states of the atom must have an infinite life-time. -8 But the excited states of the atom have life–time ~ 10 sec. The finite life-time Δt gives a finite width (uncertainty) to the energy levels. Two-slit Interference Experiment Slit 1 meter Laser Slit Source Detector Rate of photon arrival = 2 x 10 /sec Rate of photon detection = 10 /sec Time lag = 0.5 x 10 sec Spatial separation between photons = 0.5 x 10 c = 150 m – Taylor’s experiment (1908): double slit experiment with very dim light: interference pattern emerged after waiting for few weeks – interference cannot be due to interaction between photons, i.e. cannot be outcome of destructive or constructive combination of photons interference pattern is due to some inherent property of each photon - it “interferes with itself” while passing from source to screen – photons don’t “split” – light detectors always show signals of same intensity – slits open alternatingly: get two overlapping single-slit diffraction patterns – no two-slit interference – add detector to determine through which slit photon goes: no interference – interference pattern only appears when experiment provides no means of determining through which slit photon passes Double slit experiment – QM interpretation – patterns on screen are result of distribution of photons – no way of anticipating where particular photon will strike – impossible to tell which path photon took – cannot assign specific trajectory to photon – cannot suppose that half went through one slit and half through other – can only predict how photons will be distributed on screen (or over detector(s)) – interference and diffraction are statistical phenomena associated with probability that, in a given experimental setup, a photon will strike a certain point – high probability bright fringes – low probability dark fringes Double slit expt. -- wave vs quantum wave theory quantum theory • pattern of fringes: • pattern of fringes: – Intensity bands due to – Intensity bands due to variations in square of variations in probability, P, amplitude, A , of resultant of a photon striking points wave on each point on on screen screen • role of the slits: • role of the slits: – to provide two coherent – to present two potential sources of the secondary routes by which photon can waves that interfere on the pass from source to screen screen Wave function The quantity with which Quantum Mechanics is concerned is the wave function of a body. Wave function, ψ is a quantity associated with a moving particle. It is a complex quantity. 2 |Ψ| is proportional to the probability of finding a particle at a particular point at a particular time. It is the probability density. 2 | | * ψ is the probability amplitude. Thus if AiB then * AiB 2 2 2 2 2 2 | | * A i B A B Normalization |Ψ| is the probability density. The probability of finding the particle within an element d volume 2 | | d Since the particle is definitely be somewhere, so 2 | | d 1 Normalization A wave function that obeys this equation is said to be normalized. Properties of wave function 1. It must be finite everywhere. If ψ is infinite for a particular point, it mean an infinite large probability of finding the particles at that point. 2. It must be single valued. If ψ has more than one value at any point, it mean more than one value of probability of finding the particle at that point which is obviously ridiculous. 3. It must be continuous and have a continuous first derivative everywhere. , , must be continuous x y z 4. It must be normalizable. Schrodinger’s time independent wave equation One dimensional wave equation for the waves associated with a moving particle is Asin 2x (i) where ψ is the wave amplitude for a given x. A is the maximum amplitude. λ is the wavelength From (i) 42 2x 2 2 Asin x 2 2 4 (ii) x 2 2 h m v o 1 2 2 2 2m o mov 1 mov 2 2 h 2 h2 1 2m o 2 2 h where K is the K.E. for the non-relativistic case Suppose E is the total energy of the particle and V is the potential energy of the particle 1 2m o 2 2 (E V) (iii) h Equation (ii) now becomes 2 4 2 2m oE V) x 2 h2 2 2m o 2 2 (E V) 0 x This is the time independent (steady state) Schrodinger’s wave equation for a particle of mass m , total energy E, potential o energy V, moving along the x-axis. If the particle is moving in 3-dimensional space then 2 2 2 2m o 2 2 2 2 (E V) 0 x y z 2 2m o 2 (E V) 0 This is the time independent (steady state) Schrodinger’s wave equation for a particle in 3-dimensional space. For a free particle V = 0, so the Schrodinger equation for a free particle 2 2m o 2 E 0 Schrodinger’s time dependent wave equation Wave equation for a free particle moving in +x direction is i Ae (Etpx) (i) where E is the total energy and p is the momentum of the particle Differentiating (i) twice w.r.t. x 2 2 2 p p2 2 (ii) x 2 2 x2 Differentiating (i) w.r.t. t iE E i (iii) t t For non-relativistic case E = K.E. + Potential Energy p2 E V x,t 2m 2 E p V (iv) 2m Using (ii) and (iii) in (iv) 2 2 i t 2m x2 V This is the time dependent Schrodinger’s wave equation for a particle in one dimension. Linearity and Superposition If ψ1and ψ ar2 two solutions of any Schrodinger equation of a system, then linear combination of ψ a1d ψ wil2 also be a solution of the equation.. a 1a1 2 2 is also a solution Here a &a are constants 1 2 Above equation suggests: (i) The linear property of Schrodinger equation (ii) ψ and ψ follow the superposition principle 1 2 If P is the probability density corresponding to ψ and P is the 1 1 2 probability density corresponding to ψ 2 Then 1 2 due to superposition principle Total probability will be 2 2 P | | | 1| 2 * ( 1 ) ( 2 1 ) 2 * * ( 1 )( 2 1 ) 2 * * * * 1 1 2 2 1 2 2 1 * * P P 1P 2 1 2 2 1 P P 1P 2 Probability density can’t be added linearly Expectation values Expectation value is the value of ‘x’ we would obtain if we have measured the positions of a large number of particles described by the same function at some instant ‘t’ and then averaged the results. Expectation value of any quantity which is a function of ‘x’ ,say f(x) is given by f (x) f (x)| | dx for normalized ψ Thus expectation value for position ‘x’ is x x| | dx Q. Find the expectation value of position of a particle having wave function ψ = ax between x = 0 & 1, ψ = 0 elsewhere. Solution 1 1 x x| | dx2 a2 x dx 0 0 4 1 2x a 4 0 2 a x 4 Operators (Another way of finding the expectation value) An operator is a rule by means of which, from a given function we can find another function. For a free particle (Etpx) Ae Then i p x Here ^ p (i) i x is called the momentum operator Similarly i E t Here ^ E i (ii) t is called the Total Energy operator Equation (i) and (ii) are general results and their validity is the same as that of the Schrodinger equation. If a particle is not free then ^ ^ ^ ^ ^ p 2 ^ E K.E.U E U 2m o 2 ^ 1 U U U t 2m x 2 2 i 2U t 2m x 2 2 i U t 2m x 2 This is the time dependent Schrodinger equation If Operator is Hamiltonian ^ 2 2 H U 2m x 2 Then time dependent Schrodinger equation can be written as ^ H E This is time dependent Schrodinger equation in Hamiltonian form. Eigen values and Eigen function Schrodinger equation can be solved for some specific values of energy i.e. Energy Quantization. The energy values for which Schrodinger equation can be solved are called ‘Eigen values’ and the corresponding wave function are called ‘Eigen function’. Suppose a wave function (ψ) is operated by an operator ‘β’ such that the result is the product of a constant saya’ and the wave function itself i.e. ^ a then ^ ψ is the eigen function of ^ a is the eigen value of 2 2x d Q. Suppose e is eigen function of operator2 then find the eigen value. dx Solution. ^ d 2 G 2 dx ^ 2 2 G d d (e ) dx2 dx2 ^ 2x G 4e ^ G 4 The eigen value is 4. Particle in a Box Consider a particle of rest mass m enclosed in a one-dimensional box (infinite potential well). V V Boundary conditions for Potential 0 for 0 < x < L V(x)={ particle for 0 >= x >= L Boundary conditions for ψ V 0 0 for x = 0 Ψ = { x = 0 x = L 0 for x = L Thus for a particle inside the box Schrodinger equation is 2m o 2 2 E 0 (i) V 0 inside x h 2 (k is the propagation constant) p k k p 2moE 2m E k2 o (ii) 2 Equation (i) becomes 2 2 x 2k 0 (iii) General solution of equation (iii) is (x) Asin kx Bcos kx (iv) Boundary condition says ψ = 0 when x = 0 (0) Asin k.0 Bcos k.0 0 0 B.1 B 0 Equation (iv) reduces to (x) Asin kx (v) Boundary condition says ψ = 0 when x = L (L) Asin k.L 0 Asin k.L A 0 sin k.L 0 sin k.L sin n kL n n k (vi) L Put this in Equation (v) n x ( ) Asin L When n # 0 i.e. n = 1, 2, 3…., this gives ψ = 0 everywhere. Put value of k from (vi) in (ii) k 2moE 2 2 n 2m o 2 L 2 2 k 2 n h E 2 (vii) 2m o 8m o Where n = 1, 2, 3…. Equation (vii) concludes 1. Energy of the particle inside the box can’t be equal to zero. The minimum energy of the particle is obtained for n = 1 2 E h (Zero Point Energy) 1 8m L 2 o IfE 0 momentum 0 i.e.p 0 1 x Butx max L since the particle is confined in the box of dimension L. Thus zero value of zero point energy violates the Heisenberg’s uncertainty principle and hence zero value is not acceptable. 2. All the energy values are not possible for a particle in potential well. Energy is Quantized 3. E nre the eigen values and ‘n’ is the quantum number. 4. Energy levels (E ) are not equally spaced. n n = 3 E 3 n = 2 E 2 n = 1 E 1 x( ) A sinn x n L Using Normalization condition | (x)| dx 1 n L 2 2n x A sin dx 1 0 L 2 A 1 A L The normalized eigen function of the particle are 2 nx nx) sin L L Probability density figure suggest that: 1. There are some positions (nodes) in the box that will never be occupied by the particle. 2. For different energy levels the points of maximum probability are found at different positions in the box. |ψ 1 is maximum at L/2 (middle of the box) 2 |ψ 2 is zero at L/2. Particle in a Three Dimensional Box Eigen energy E E E E x y z h2 E (nx n y nz) 8mL 2 Eigen function x y z n x nyy n z AxAyA zsin x sin sin z L L L 3 2 nx n y nz L sin L sin L sin L Statistical Mechanics Statistical Distribution This determines the most probable way in which a certain total amount of energy ‘E’ is distributed among the ‘N’ members of a system of particles in thermal equilibrium at absolute temperature, T. Thus Statistical Mechanics reflects overall behavior of system of many particles. Suppose n(є) is the no. of particles having energy, є then n() g() f () where g(є) = no. of states of energy є or statistical weight corresponding to energy є f(є) = distribution function = average no. of particles in each state of energy є = probability of occupancy of each state of energy є Statistical Distribution are of three kinds 1. Identical particles that are sufficiently far apart to be distinguishable. Example: molecules of a gas. Negligible overlapping of ψ Maxwell-Boltzmann Statistics 2. Indistinguishable identical particles of ‘0’ or integral spin. Example: Bosons (don’t obey the exclusion principle) Overlapping of ψ Bose-Einstein Statistics 3. Indistinguishable identical particles with odd-half integral spin (1/2, 3/2, 5/2 ..). Example: Fermions (obey the exclusion principle) Fermi-Dirac Statistics Maxwell-Boltzmann Statistics (Classical Approach) According to this law number of identical and distinguishable particles in a system at temperature, T having energy є is n(є) = (No. of states of energy є).(average no. of particles in a state of energy є) n ) g ) Ae kT (i) kT Here A is a constant and fM . . ) Ae Equation (i) represents the Maxwell-Boltzmann Distribution Law This law cannot explain the behavior of photons (Black body radiation) or of electrons in metals (specific heat, conductivity) Applications of M.B. Statistics (i) Molecular energies in ideal gas No. of molecules having energies between є and є + dє is given by n()d {g()d}{f ()} kT (i) n()d Ag()e d The momentum of a molecule having energy є is given by md p 2m dp 2m No. of states in momentum space having momentum between p and p + dp is proportional to the volume element i.e. dpxdp yp z 4 3 4 3 3 (p dp ) 3 p 4p dp g(p)dp 2 4p dp g(p)dp Bp dp B is a constant Each momentum corresponds to a single є 2 g()d Bp dp md g( )d B m . 2 m 3/2 g()d 2m B d Put in (i) No. of molecules with energy between є and є + dє n()d 2m 3/2BA e kTd kT n()d C e d (ii) Total No. of molecules is N kT N n()d C e d 0 0 Using identity 1 ax 1 x e dx 0 2a a C 2N (kT) 32 Put value of C in equation (ii) 2N kT n()d 2 e d ( kT) This equation gives the number of molecules with energies between є and є + dє in a sample of an ideal gas that contains N molecules and whose absolute temperature is T. It is called molecular energy distribution equation. n(є) 0 kT 2kT 3kT Maxwell-Boltzmann energy distribution curve for the molecules of an ideal gas. Average Molecular Energy Total energy of the systE ()d 0 2N 3 E 32 e kTd (kT ) 0 Using x e axdx 3 4a 2 a 0 2N 3 E 3 (kT)2 kT (kT ) 2 4 3 E 2 NkT E Average energy per mol e N 3 kT 2 This is the average molecular energy and is independent of the molecular mass At Room Temperature its value is 0.04 eV. Equipartition of Energy: Equipartition Theorem Average Kinetic Energy associated with each degree of freedom of a molecule is1 kT 2 Distribution of molecular speed 1 2 K.E. mv 2 d mvdv Put this in molecular energy distribution equation 3 m 2 mv n ( )v 4N v e 2kdv 2kT It is called molecular speed distribution formula. RMS Speed 2 vrms v It is the square root of the average squared molecular speed 1 2 3 2 mv 2 kT 3kT vrms m Average Speed 1 v vn(v)dv N 0 32 m 3 v22kT 4 2kT v e dv 0 Using 3 ax2 1 x e dx 2 0 2a m 324 k T 2 v 4 . 8kT 2kT 2 m 2 m Most Probable Speed 3 m 2 2 mv n ( ) N v e 2kT 2kT 3 dn(v) m 2d 2 m2kT 4 N v e 0 dv 2 kT dv 2 m 4 N 2kT 0 d v e m2kT 0 v 2kT dv p m vmost probablp v n(v) vrms v M.B. Speed Distribution Curve (i) Speed distribution is not symmetrical (ii) vp v vrms 2kT 8kT 3kT m m m Variation in molecular speed with ‘T’ and ‘molecular mass’ This curve suggests that most probable speed α T Also the most probable speed α 1/m Quantum Statistics (Indistinguishable identical particles) Bosons Fermions 1. ‘0’ or integral spin 1. Odd half integral spin 2. Do not obey exclusion 2. Obey exclusion principle principle 3. Symmetric wave function 3. Anti-symmetric wave function 4. Any number of bosons 4. Only one fermion can exist in the same can exist in a particular quantum state of the quantum state of the system system Consider a system of two particles, 1 and 2, one of which is in state ‘a’ and the other in state ‘b’. When particles are distinguishable then ' (a) b2) " (2a b1) When particles are indistinguishable then For Bosons, 1 B a1) b2) (2a b1) 2 symmetric 1 For Fermions, F a1) b2) (2a b1) 2 anti-symmetric ψFbecomes zero when ‘a’ is replaced with ‘b’ i.e. 1 F a1) a2) (2a (a) 0 2 Thus two Fermions can’t exist in same state (Obey Exclusion Principle) Bose-Einstein Statistics Any number of particles can exist in one quantum state. Distribution function can be given by 1 fB. . ) e e kT1 where α may be a function of temperature, T. For Photon gas 0 e 1 -1 in denominator indicates multiple occupancy of an energy state by Bosons Bose-Einstein Condensation If the temperature of any gas is reduced, the wave packets grow larger as the atoms lose momentum according to uncertainty principle. When the gas becomes very cold, the dimensions of the wave packets exceeds the average atomic spacing resulting into overlapping of the wave packets. If the atoms are bosons, eventually, all the atoms fall into the lowest possible energy state resulting into a single wave packet. This is called Bose-Einstein condensation. The atoms in such a Bose-Einstein condensate are barely moving, are indistinguishable, and form one entity – a superatom. Electromagnetic Spectrum visible microwaves infrared light ultraviolet x-rays 1000 100 10 1 0.1 0.01 Low High Energy Energy (m) Black Body Radiation All objects radiate electromagnetic energy continuously regardless of their temperatures. Though which frequencies predominate depends on the temperature. At room temperature most of the radiation is in infrared part of the spectrum and hence is invisible. The ability of a body to radiate is closely related to its ability to absorb radiation. A body at a constant temperature is in thermal equilibrium with its surroundings and must absorb energy from them at the same rate as it emits energy. A perfectly black body is the one which absorbs completely all the radiation, of whatever wavelength, incident on it. Since it neither transmits any radiation, it appears black whatever the color of the incident radiation may be. There is no surface available in practice which will absorb all the radiation falling on it. The cavity walls are constantly emitting and absorbing radiation, and this radiation is known as black body radiation. Characteristics of Black Body Radiation (i) The total energy emitted per second per unit area (radiancy E or area under curve) increases rapidly with increasing temperature. (ii) At a particular temperature, the spectral radiancy is maximum at a particular frequency. (iii) The frequency (or wavelength, λ ) m for maximum spectral radiancy decreases in direct proportion to the increase in temperature. This is called “Wien’s displacement law” mT constant (2.898 x 10 m.K) Planck’s Radiation Law Planck assumed that the atoms of the walls of cavity radiator behave as oscillators with energy n nhv n 0,1,2,3.... The average energy of an oscillator is N is total no. of Oscillators N hv hv (i) e kT1 Thus the energy density (uv) of radiation in the frequency range v to v + dv is 2 8v dv uvdv 3 c 8 v dv hv uvdv 3 hv c e kT1 8 hv 3 dv u vv 3 hkT c e 1 This is Planck’s Radiation formula in terms of frequency. In terms of wavelength 8 hc d ud 5 hkT e 1 This is Planck’s Radiation formula in terms of wavelength. Case I : (Rayleigh-Jeans Law) 8 hc d ud 5 hc e k1 When λ is large then hkT hc e 1 kT u d 8 hc d 5 hc 1 kT 8 kT This is Rayleigh-Jeans ud 4 d Law for longer λ’s. Case II : (Wein’s Law) u d 8 hc d 5 e hk 1 When λ is very small theehkT1 8hc hc u d e kd 5 This is Wein’s Law for small λ’s. A B This is another form of ud 5e kd Wein’s Law.. Here A and B are constants. Stefan’s Law Planck’s Radiation formula in terms of frequency is 3 8 hv dv u vv 3 hkT c e 1 3 cu dv 2 hv dv 4 v c2 ehvkT1 The spectral radiancyvE is related to the energy devsity u by c Ev uv 4 3 2 h v dv E vv 2 hkT c e 1 2h v dv E Edvv 2 hv 0 c 0 e kT1 hv kT kT Let x v x and dv dx kT h h 4 4 3 2 k T x dx E 3 2 x h c 0e 1 3 4 But x dx e 1 15 0 2 5k4 E T 4 15h c 2 5 4 2 k Let 3 2 (Stefan’s constant) 15h c 4 E T This is Stefan’s Law. 2 5k4 Here 3 2 = 5.67 x 10 erg/(cm .sec.K ) 15h c = 5.67 x 10 W/(m .K ) Wein’s Displacement Law Planck’s Radiation formula in terms of wavelength is 8 c 1 u hc 5 e k1 d u ) Using 0 for max d hc Constant kT max mT constant (2.898 x 10 m.K) Peaks in Black Body radiation shifts to shorter wavelength with increase in temperature. Specific Heat of Solids Atoms in solid behave as oscillators. In case of solids total average energy per atom per degree of freedom is kT (0.5kT from K.E. and 0.5kT from P.E.). Each atom in the solid should have total energy = 3kT (3 degrees of freedom). For one mole of solid, total energy, Eo= 3N kT (classically) Here N os the Avogadro number. E = 3RT Specific heat at constant volume is E C v 3 R ~ 6 kcal/kmol.K T V This is Dulong & Petit’s Law. This means that atomic specific heat (at constant volume) for all solids is approx 6 kcal/kmol.K and is independent of T. Dulong and Petit’s Law fails for light elements such as B, Be and C. Also it is not applicable at low temperatures for all solids. Einstein’s theory of Specific Heat of Solids According to it the motion of the atoms in a solid is oscillatory. The average energy per oscillator atom) is hv hkT e 1 The total energy for one mole of solid in three degrees of freedom. hv E 3N o hv e kT1 E The molar specific heat of soliC v T V hv ehvkT 2 hvkT C v 3N ov hv 3N k hv hv kT 2 (e kT1) 2 o kT (e kT1) 2 2 hkT C 3R hv e v kT (e hkT1) 2 This is Einstein’s specific heat formula. Case I: At high T, kT >> hv then hv hv e kT1 kT 2 1 hv hv kT C v R3 kT 2 1 hv 1 kT hv C v 3R 1 kT Cv 3R kT hv) This is in agreement with Dulong & Petit’s Law at high T. Case II: At low T, hv >> kT thene hkT1 2 hv hkT Cv 3R e kT This implies that aT 0,C v0 i.e. is in agreement with experimental results at low T. Fermi-Dirac Statistics Obey Pauli’s exclusion Principle Distribution function can be given by f () 1 F.D. e e kT1 f (є) can never exceed 1, whatever be the value of α, є and T. So only one particle can exist in one quantum state. α is given by F kT where F is the Fermi Energy then 1 f ( ) F D. (F kT e 1 Case I: T = 0 ( ) kT For є < є F F ( ) kT For є > є F F For є < F 1 f () 1 F.D. e 1 For є > F f ( ) 1 0 F. . e 1 Thus at T = 0, For є < є aFl energy states from є = 0 to є F are occupied as fF.D.() 1 Thus at T = 0, all energy states for which є > є Fre vacant. f () 0 F.D. Case II: T > 0 Some of the filled states just below є Fecomes vacant while some just above є bFcome occupied. Case III: At є = F 1 1 fF D . ) For all T e 1 2 Probability of finding a fermion (i.e. electron in metal) having energy equal to fermi energy Fє is ½ at any temperature. fF D.() 0 K 1 0.5 10000 K 1000 K 0 1 єF Є (eV) Free electrons in a metal Typically, one metal atom gives one electron. One mole atoms gives one mole of free electoons, (N ) If each free electron can behave like molecules of an ideal gas. Then E 3 N kT 3 RT Average K.E. for 1 mole of electroe g2s,o 2 Ee 3 Molar specific heat of electrCvges R T V 2 Then total specific heat in metals at high T should be 3 9 Cv 3R R R 2 2 But experimentally at high TC 3R v Free electrons don’t contribute in specific heat, Why? Electrons are fermions and have upper limit on the occupancy of the quantum state. By definition highest state of energy to be filled by a free electron at T = 0 is obtained aF є = є The no. of electrons having energy є is 3 F 8 2 m 2 N g d) where g ()d 3 d 0 h Here V is the volume of the metal 32F 2 N 8 2 Vm d 16 2 Vm 32 h3 3h 3 F 0 2 23 h 3N F (i) 2 m 8 V where N / V is the density of free electrons. Electron energy distribution No. of electrons in the electron gas having energy between є and є + dє is n()d g() f ()d 8 2 V 2 32 n )d 3 ( )/kT d h e F 1 using (i) 2 n d 3 N d 2 e(F )/k1 This is electron energy distribution formula, according to which distribution of electrons can be found at different temperatures. When a metal is heated then only those electrons which are near the top of the fermi level (kT of the Fermi energy) are excited to higher vacant energy states. kT = 0.0025 eV at 300K kT = 0.043 eV at 500K The electrons in lower energy states cannot absorb more energy because the states above them are already filled. This is why the free electrons contribution in specific heat is negligible even at high T. Average electron energy at 0K Total energy at 0K is F E o n()d 0 Since at 0K all electrons have energy less than or equal to є F (F)/T e e 0 F 3N 32 32 3N E o F d F 2 0 5 Then average electron energy E 3 o o F at 0K. N 5

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