Rohlf Chapter 12, #10,13,22

Physical constants
Boltzmann's constant  k= 1.38066E-23 J/K
k= 0.00008617 eV/K
Atomic mass unit   u= 1.66E-27 kg
umev= 931.5 MeV/c^2
Wien constant   2.90E-03
Planck's constant  h= 4.14E-15 eV s
h= 6.63E-34 J s
h/ 2 pi hbar 1.05E-34 J s
h/ 2 pi hbev 6.59E-16 eV s
Speed of light  c= 3.00E+08 m/s
hc= 1.24E+03 eV nm
hbarc= 1.97E+02 eV nm
Electron charge  e= 1.60E-19 C
electron volt  eV= 1.60E-19 joule
electron mass  mel 9.11E-31 kg
mel 5.11E-01 MeV/c^2
proton mass  mp 1.67E-27 kg
mpev 9.38E+02 MeV/c^2
mpg 0.93827231 GeV/c^2
neutron mass  mn 1.67E-27 kg
mnev 9.40E+02 MeV/c^2
mng 0.93956563 GeV/c^2
Stefan's constant=  sigma= 5.67E-08 watt/m^2K^4
Nuclear distance unit  fermi 1.00E-15 m (a femtometer)
cross section unit  barn= 1.00E-28 m^2
barn= 100 fm^2
common combo for scattering  ke^2 1.44 MeV fm

First Bohr orbit radius a0  a0 0.0529 nm
Bohr magnetion  mub 5.79E-05 eV/T
muba 9.27E-24 A m^2 A=ampere

#10 Distributions of 5 particles sharing 10 units of energy

The following table shows the 30 possible distributions (macrostates). For Maxwell-Boltzmann statistics, the particles are distinguishable and there are a number of microstates based on the number of different ways to distribute the particles in each macrostate. The number of microstates is listed in the last column. There is a total of 1001 microstates for the Maxwell-Boltzmann distribution, but only the 30 states for the Einstein-Bose case where the particles are indistinguishable.

 ... 0 1 2 3 4 5 6 7 8 9 10 Microstates 1 4 0 0 0 0 0 0 0 0 0 1 5 2 3 1 0 0 0 0 0 0 0 1 0 20 3 3 0 1 0 0 0 0 0 1 0 0 20 4 3 0 0 1 0 0 0 1 0 0 0 20 5 3 0 0 0 1 0 1 0 0 0 0 20 6 3 0 0 0 0 2 0 0 0 0 0 10 7 2 2 0 0 0 0 0 0 1 0 0 30 8 2 1 1 0 0 0 0 1 0 0 0 60 9 2 1 0 1 0 0 1 0 0 0 0 60 10 2 1 0 0 1 1 0 0 0 0 0 60 11 2 0 2 0 0 0 1 0 0 0 0 30 12 2 0 1 1 0 1 0 0 0 0 0 60 13 2 0 1 0 2 0 0 0 0 0 0 30 14 2 0 0 2 1 0 0 0 0 0 0 30 15 1 3 0 0 0 0 0 1 0 0 0 20 16 1 0 3 0 1 0 0 0 0 0 0 20 16 1 2 1 0 0 0 1 0 0 0 0 60 16 1 1 2 0 0 1 0 0 0 0 0 60 17 1 2 0 1 0 1 0 0 0 0 0 60 18 1 2 0 0 2 0 0 0 0 0 0 30 19 1 1 1 1 1 0 0 0 0 0 0 120 20 1 1 0 3 0 0 0 0 0 0 0 20 21 1 0 2 2 0 0 0 0 0 0 0 30 22 0 4 0 0 0 0 1 0 0 0 0 5 23 0 3 1 0 0 1 0 0 0 0 0 20 24 0 3 0 1 1 0 0 0 0 0 0 20 25 0 2 2 0 1 0 0 0 0 0 0 30 26 0 2 1 2 0 0 0 0 0 0 0 30 27 0 1 3 1 0 0 0 0 0 0 0 20 28 0 0 5 0 0 0 0 0 0 0 0 5 Sum 44 33 27 16 11 7 5 3 2 1 1 1001

The average number of particles per energy state can be calculated for each type of statistics. These averages are shown in the table below.

 Energy BoltzmannAvg population Einstein-BoseAvg population Fermi-DiracAvg population 0 1.429 1.467 1.375 1 1.099 1.100 1.063 2 0.824 0.900 0.875 3 0.599 0.533 0.625 4 0.420 0.367 0.500 5 0.280 0.233 0.250 6 0.175 0.167 0.188 7 0.100 0.100 0.063 8 0.050 0.067 0.063 9 0.020 0.033 0 10 0.005 0.033 0

The states possible with Fermi-Dirac statistics are shown below. It is assumed that we are dealing with electrons with two spin states, so each state can have up to two particles. These 16 allowed states are a subset of the 30 states which were possible with Maxwell-Boltzmann and Bose-Einstein statistics which allow an unlimited population in any given state.

 ... 0 1 2 3 4 5 6 7 8 9 10 1 2 2 0 0 0 0 0 0 1 0 0 2 2 1 1 0 0 0 0 1 0 0 0 3 2 1 0 1 0 0 1 0 0 0 0 4 2 1 0 0 1 1 0 0 0 0 0 5 2 0 2 0 0 0 1 0 0 0 0 6 2 0 1 1 0 1 0 0 0 0 0 7 2 0 1 0 2 0 0 0 0 0 0 8 2 0 0 2 1 0 0 0 0 0 0 9 1 2 1 0 0 0 1 0 0 0 0 10 1 2 0 1 0 1 0 0 0 0 0 11 1 2 0 0 2 0 0 0 0 0 0 12 1 1 1 1 1 0 0 0 0 0 0 13 1 0 2 2 0 0 0 0 0 0 0 14 1 1 2 0 0 1 0 0 0 0 0 15 0 2 2 0 1 0 0 0 0 0 0 16 0 2 1 2 0 0 0 0 0 0 0 Sum by state 22 17 14 10 8 4 3 1 1 0 0

#13. You can argue for the electrons simply on the basis that their DeBroglie wavelengths
are much smaller, but try a look at the density of states.
For electrons the density is = 4*pi*(2m)^(3/2)*sqrt(E)/h^3 =
= 4*pi*(2*mc^2)^(3/2)*sqrt(E))/(hc)^3 =
= 4*pi*(2*.511E6 eV)^(3/2)*sqrt(1 eV))/(1240 eV nm)^3 =
6.81/nm^3

So for electrons the density is about 6.8 per cubic nanometer.

For photons the density is 8 pi e^2/(hc)^3 = 8 pi /(1240 eV nm)^3 =    1.31818E-08 1/nm^3

So the photon density under present conditions is 8 orders of magnitude smaller.

#22.  Quantum or classical statistics for the neutron beam?
Neutron flux 10^13 /m^2 s at 300K.
Take their kinetic energy to be 3kT/2 and calculate velocity
At 300K, the 3kT/2 energy is   0.038 eV
This is non relativistic, so mv^2/2 can be used to calculate the velocity.
Using neutron mass of   940 MeV
v=3E8*sqrt(2*.038 eV/9.4E8 eV) =   2697.516588 m/s
Considering the beam to be a cylinder of area 1 m^2 and length 2700 meters, we can calculate density.
density = 10^13/m^2 s/2697.5 m/s =    3707136237  /m^3
Taking the separation to be 1/cube root of density, separation =    0.000646 m
The DeBroglie wavelength for the neutron is = h/mv = hc*c/mc^2 v =    0.1466 nm
So the separation is very large compared to the DeBroglie wavelength, making them distinguishable
and subject to Maxwell-Boltzmann statistics.