Lecture_24 - PHYS342 Fall2011 Lecture 24: Boltzmann...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
PHYS 342 Fall 2011 Lecture 24: Boltzmann Equation, Number of Available States, Detailed Statistics for a Few Particles with Quantized Energies Ron Reifenberger irck anotechnology Center Birck Nanotechnology Center Purdue University 1 Lecture 24 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
The Maxwell-Boltzmann Factor: The System Inputs: N=4 distinguishable particles (to illustrate the ideas) Equally spaced allowed energies 0 ε , 1 ε , 2 ε …. The allowed energies are now QUANTIZED. Add fixed amount of energy, say E tot =5 ε (to illustrate the ideas) Let n =number of particles Ln 1 num f p with energy ε 1 , etc. How many available states oes the system have for a 5 ε does the system have for a fixed total energy of 5 ε ? 12 6 ( , ,...... ) 5 tot r r r r En n n n 2 2 4 r r Nn 
Background image of page 2
State A : B : C : Possible States for a Fixed Energy of E tot =5 ε 4 ε 5 ε 4 ε 5 ε 3 ε 2 ε 3 ε 2 ε Microscopic Model 1 ε 0 ε 1 ε 0 ε State D : E : F : 4 ε 5 ε 4 ε 5 ε 3 ε 2 ε 3 ε 2 ε 3 1 ε 0 ε 1 ε 0 ε
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
How many microstates for each macrostate (E tot =5 ε )? Equally Spaced Energy Levels; 4 Distinguishable Particles; Total Energy FIXED Particles are distinguishable:  ! !! N N n nN n    4 4 Macro state 0      TOTAL E Microstates g(E) A3 0 0 0 0 1 5 4 2 1 43 12 11 3 B 2 1 0 0 1 0 5 12 C2 0 1 1 0 0 5 12 D1 2 0 1 0 0 5 12 E1 1 2 0 0 0 5 12 12 12 F 0 3 1 0 0 0 5 4 56 .0 ( 0 ) Prob that one particle has P 4 1 21 2 4 2 1 4 56 4 56 4 56 4 56 4 32211 456 56 0 56   1( 1 ) rob that one particle has P . 4 1 2 4 1 5 4 56 4 56 4 56 4 56 4 01021 3 56 Prob that one particle has  
Background image of page 4
Situation Symbol Probability ti l 0 (0 1/56 Probability of finding a particle in a given energy state particle w. 0 ε P(0 ε ) 21/56 particle w. 1 ε P(1 ε ) 15/56 particle w. 2 ε P(2 ε ) 10/56 particle w. 3 ε P(3 ε )6 / 5 6 particle w. 4 ε P(4 ε )3 / 5 6 article w. 5 (5 /56 particle w. 5 ε P(5 ε ) 1/56 TOTAL 56/56 g tl 5 30 0.40 0.50 of findin energy E E total =5 ε Exponential fit: E e 00 0.10 0.20 0.30 obabilty o article w. 5 0.00 012345 Pr pa Energy(units of ε )
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Why is an exponential a good fit? -a plausibility argument - 1. Maxwell-Boltzmann distribution for gas atoms 2. Naturally arises for N particles (N must be large) where Probability that one particle has energy E 1 = P(E 1 ) Probability that another particle has energy E 2 = P(E 2 ) robability that particle one has energy E ND particle two has energy the sum of all energies is some constant E tot .
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/08/2011 for the course PHYS 342 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

Page1 / 22

Lecture_24 - PHYS342 Fall2011 Lecture 24: Boltzmann...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online