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P460 - Quan. Stats. II1 Quantum Statistics:Applications Determine P(E) = D(E) x n(E) probability(E) = density of states x prob. per state electron in Hydrogen.

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Presentation on theme: "P460 - Quan. Stats. II1 Quantum Statistics:Applications Determine P(E) = D(E) x n(E) probability(E) = density of states x prob. per state electron in Hydrogen."— Presentation transcript:

1 P460 - Quan. Stats. II1 Quantum Statistics:Applications Determine P(E) = D(E) x n(E) probability(E) = density of states x prob. per state electron in Hydrogen atom. What is the relative probaility to be in the n=1 vs n=2 level? D=2 for n=1 D=8 for n=2 as density of electrons is low can use Boltzman: can determine relative probability If want the ratio of number in 2S+2P to 1S to be.1 you need T = 32,000 degrees. (measuring the relative intensity of absorption lines in a star’s atmosphere or a interstellar gas cloud gives T)

2 P460 - Quan. Stats. II2 1D Harmonic Oscillator Equally spaced energy levels. Number of states at each is 2s+1. Assume s=0 and so 1 state/energy level Density of states N = total number of “objects” (particles) gives normalization factor for n(E) note dependence on N and T

3 P460 - Quan. Stats. II3 1D H.O. : BE and FD Do same for Bose-Einstein and Fermi-Dirac “normalization” varies with T. Fermi-Dirac easier to generalize T=0 all lower states fill up to Fermi Energy In materials, E F tends to vary slowly with energy (see B FD for terms). Determining at T=0 often “easy” and is often used. Always where n(E)=1/2

4 P460 - Quan. Stats. II4 Density of States “Gases” # of available states (“nodes”) for any wavelength wavelength --> momentum --> energy “standing wave” counting often holds:often called “gas” but can be solid/liquid. Solve Scrd. Eq. In 1D go to 3D. n i >0 and look at 1/8 of sphere 0 L

5 P460 - Quan. Stats. II5 Density of States II The degenracy is usually 2s+1 where s=spin. But photons have only 2 polarization states (as m=0) convert to momentum convert to energy depends on kinematics relativistic non-realtivistic

6 P460 - Quan. Stats. II6 Plank Blackbody Radiation Photon gas - spin 1 Bosons - dervied from just stat. Mech. (and not for a particular case) by S.N. Bose in 1924 Probability(E)=no. photons(E) = P(E) = D(E)*n(E) density of state = D(E) = # quantum states per energy interval = n(E) = probability per quantum state. Normalization: number of photons isn’t fixed and so a single higher E can convert to many lower E energy per volume per energy interval =

7 P460 - Quan. Stats. II7 Phonon Gas and Heat Capacity Heat capacity of a solid depends on vibrational modes of atoms electron’s energy levels forced high by Pauli Ex. And so do not contribute most naturally explained using phonons - spin 1 psuedoparticles - correspond to each vibrational node - velocity depends on material acoustical wave EM wave phonon photon almost identical statistical treatment as photons in Plank distribution. Use Bose statistics done in E&R Sect 11-5, determines


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