Quantum Physics Lecture 5

Transcription

1 Quantum Physics Lecture 5 Thermal Phenomena - continued Black Body radiation - Classical model, UV catastrophe - Planck model, Wien & Stefan laws - Photoelectric effect revisited The hydrogen atom Planetary model Energy and orbit radius Bohr model

2 Blackbody Radiation At finite temperature matter glows i.e. emits radiation with a continuous spectrum. e.g Infrared imaging of people, planet etc. Surface dependent (emissivity, silvery, black, etc.) Blackbody = ideal 100% emitter/absorber in thermal equilibrium with its surroundings. Practical realisation is a thermal cavity. Measure: spectrum energy density u(ω) Observe that increasing temperature (1) increases u overall (2) shifts peak emission to higher frequencies i.e. colour and intensity of hot objects vary with T Examples Bar fire, molten iron, stars, universe µwave background..

3 Rayleigh-Jeans model Classical cavity walls in thermal equilibrium with waves inside. Assumptions: (1) Oscillators in walls emit/absorb EM waves (2) Walls are perfect reflectors standing waves (3) Equipartition of Energy i.e. < E > = k b T/2 per degree of freedom, So k b T per oscillator/wave Problem to solve is How many waves/oscillators, and at what wavelengths?

4 Waves in boxes: 1-D & 3-D 1-D box size L Standing waves with 2L λ = j where j = 1,2,3. For 3-D: cube of side L j x, j y, j z = 1,2,3 independently for each axis Standing waves, nodes at walls, modes of cavity Wave equation solution shows j 2 = j x 2 + j y 2 + j z 2 Next question: How many such waves in cavity have wavelengths between λ and λ+dλ? Examples in a 2-D box 2,3 and 3,2 wave patterns

5 j-space calculation j values of possible waves are a cubic grid of points with separation 2L/λ values 1,2,3 etc. on each axis This is j-space So g(j)dj - the number of waves in range λ to λ+dλ is the number of points in j-space in a spherical shell radii j to j+dj Assume j large, so j is (almost) continuous so g( j)dj = 4πj 2 dj 5 Relate g(λ) to g(j) Need 2 modifications: λ is +ve octant only, so reduce by factor 8 2-polarisation modes, so increase by x2 Overall, reduce by x j j+dj g(λ)dλ = 1 4 g( j)dj = πj 2 dj

6 Number of waves & energy So ( )dλ = π j 2 dj Now j = 2L g λ g( λ)dλ = π 2L 2 2L λ λ dλ = 8π L 2 λ 4 3 λ dλ dj = 2L λ 2 dλ Note λ -4 dependence Energy Energy density in cavity Convert to frequency E( λ)dλ = k b T 8π L3 λ 4 dλ u( λ)dλ = 1 L k 8π L3 3 bt λ = c f = 2πc ω u( ω )dω = 8πk Tω 4 b 2π ( ) 4 c 4 2πc dλ = 2πc ω dω 2 λ 4 ω dω = k Tω 2 b dω 2 π 2 c 3 dλ = 8πk bt λ 4 dλ Classical result, diverges at high ω the Ultraviolet catastrophe!

7 Comparison with experiment? kbt ω 2 u (ω ) dω = 2 3 dω π c OK at low frequency. Fails at high frequency The U-V catastrophe (of classical physics!) u vs. ω vs. λ Empirical fit for high frequency by Wien, but no theory. The big problem for classical physics at end of C19

8 Planck s Blackbody Expression Planck: oscillators can only emit energy in packets of ħω - the original quantum hypothesis! The only possible energy levels of the oscillators are E=nħω Emission is the result of transitions between these levels. See specific heats for statistical analysis multiply classical result by ω k b T e ω k bt 1 u( ω )dω = ω 3 1 π 2 c 3 e ω kbt 1 dω Planck Blackbody radiation formula agrees with experiment First important success of quantum theory

9 Why does this quantum model work and the classical not? Classical: Any oscillator has energy k B T no matter how high frequency. High frequencies can have some amplitude and hence some energy, so the total energy diverges with the number of states and frequency. Quantum: Each oscillator exchanges energy only in discrete amounts ħω so at higher frequencies where k B T is much smaller than ħω the states are not excited and thus do not contribute to the energy. Only lower energy (longer wavelength) modes contribute. This prevents the divergence and causes intensity to fall off at higher frequencies. Analogy with specific heats. Generally, expect macroscopic quantum phenomena to be more readily visible at lower temperatures.

10 Results derived from Planck Equation - 1 Wien s Displacement Law Wavelength of maximum energy λ max 1/T (or ω max T) (i.e. more blue when hotter stars, furnace, etc.) To find ω max, differentiate Planck expression (Exercise!) d dω u ω ( )dω = 0 Find ħω max /k b T = Correct result, agrees with experiment

11 Results derived from Planck Equation - 2 Stefan Boltzmann Law: For Blackbody (perfect emitting surface) Total energy R emitted/unit area T 4 (=ST 4 ) R related to integral of Planck expression over all frequencies U = With α = ω k b T becomes u( ω )dω = ω 3 π 2 c 3 e ω kbt 1 dω 0 Integral = π 4 /15 So U = 8π 5 4 k b S-B Law! 15h 3 c T 4 = at 4 3 Can show that S = 5.67 x 10-8 W m -2 K -4 0 U = 8π k 4 b α 3 T 4 h 3 c 3 e α 1 dα 0

12 Photoelectric Effect Revisited Einstein did not solve the photoelectric effect by introducing photons to match the experiment. In fact, he predicted what the experiment should be! Confirmation of his prediction came in Actually, Einstein applied classical idea of entropy, as understood for a gas, to radiation. For a gas, ΔS = Nk b (ΔV/V) where N is number of molecules For radiation, ΔS = [E/ħω]k b (ΔV/V) [E/ħω] is number of radiation particles (Wien formula) Or ħω is the energy of light (waves) particle! An example of Quasi-history

13 Structure of the (hydrogen) atom Rutherford: small, massive +ve nucleus, surrounded by -ve electrons at (relatively) large distance. Planetary model of hydrogen: electron (e - ) orbiting proton (e + ) with speed v, at radius r e + e - Electrical force: F e = e 2 /4πε o r 2 = mv 2 /r stable orbit v = e 4πε o mr (mass centr. accel) As r decreases, v increases

14 The hydrogen atom (continued) Electron energy E = KE + PE KE (non-relativistic) = mv 2 /2 v = e 4πε o mr KE = m 2 e 4πε o mr 2 = e2 8πε o r PE (electric) = qv PE = ( e) (V: potential due to proton) e 4πε o r = e2 4πε o r E = KE + PE = e2 8πε o r e2 4πε o r = e2 8πε o r

15 The Bohr (debroglie) hydrogen atom Classically, any radius possible, therefore any energy possible! E = e2 8πε o r Fact 1: energy required to remove electron from hydrogen (eg. by photoelectric effect) is always the same E = 13.6 ev (corresponds to special radius = 5.3 x m) Fact 2: EM Theory requires an accelerating charge to radiate (emit EM waves, lose energy, causing orbit to collapse!)...should be looking for special orbits which do not radiate? (radiation to be associated with transitions between orbits)

16 The Bohr (debroglie) hydrogen atom Suggestion: An electron can circle a nucleus without losing energy if its orbit contains an integral number of de Broglie wavelengths Test? v = e 4πε o mr and λ = h mv λ = h e 4πε o r m λ = 33x10 11 m substitute r = m = 2π( m) = 2πr In general, nλ = 2πr n n=1,2,3.. See patterns in text

17 Allowed Values of Radius nλ = n h e r n = n 2 h 2 ε o πme 2 4πε o r n m = n 2 a o = 2πr n Allowed radii proportional to n 2 Bohr radius (a o ) - unfortunate label? Corresponds to n=1 Exercise: insert values of constants, show a o = x m

18 Allowed Values of Energy Allowed radii implies allowed energies E n = e2 8πε o r n = 8πε o e 2 n 2 h 2 ε o πme 2 E n = me 4 8ε o 2 h 2 1 n 2 ( ) = 2.18x10 18 ( 1 ) 2 J = 13.6( 1 ) n n 2 ev E n = E ( 1 ) 1 n 2 where E 1 = 13.6 ev Energy is proportional to 1/n 2 - compare particle in box? In this case energy is negative, -ve sign is included in E 1