Rohlf Chapter 14,#5,9,10,12,13,14,29

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

#5 Given a Fermi level of 9.39 eV for Zinc, calculate the density of conduction electrons.
Equation 14-10, p376 relates the number density to the Fermi level.
Density of electrons = n = 8*sqrt(2)*pi*m^(3/2)*((2/3)*EF^(3/2)/h^3
n = 8*sqrt(2)*pi*(mc^2)^(3/2)*((2/3)*(9.39 eV)^(3/2)/(hc)^3 =    130.625595  /nm^3
1.30626E+29  /m^3
The density of zinc is 7.13E3 kg/m^3 and its atomic mass is 65.4. The number of zinc atoms per cubic meter is
n' = (6.02E23)*(7.13E3 kg/m^3)/(65.4E-3 kg) =   6.56309E+28 atom/m^3
This gives  1.990306717 electrons per atom, and clearly indicates two conduction electrons per atom in zinc.
This is consistent with the zinc electron configuration 4s2.

#9 Given an aluminum Fermi level of 11.6 eV, calculate the Fermi speed and the electron heat capacity.
a. The Fermi speed is given by vF=c*sqrt(2*EF/mc^2) =   2021411.802 m/s
Equation 14.63, pg 384.
b. The electron contribution to the specific heat is given by eq 14.53, pg 383.
Cel = pi^2*k^2*NA*T/(2*EF)=  5.70482E+17 eV/K = 0.0913 J/K mole

#10 Find temperature where electron and phonon contributions to specific
heat are equal in copper. Given EF=7.0 eV and Debye temperature 343K
From eq. 14.53 on pg 383 we can set the contributions equal. This gives
k/(2*EF)= 12*pi^2*T^2/(5*TD^3),  so that T=sqrt(5*k*TD^3/(24*pi^2*EF)
T=sqrt(5*k*343^3/(24*pi^2*7)=   3.238 K

#12 Typical phonon and electron energies in copper at room temperature.
The typical phonon energy is given by kTD/2. Using the value of Debye temperature in #10
Phonon energy = k*343/2=  0.015 eV

The typical electron energy is given by 3*EF/5 = 3*7 eV/5 =     4.2 eV

#13 Melting point and density as indicator of Debye temperature
Figure 14-10 on page 381 shows the correlation between Debye temperature and melting point.
Given a melting point of 934K for aluminum and a density of 2.7E3 kg/m^3
Since there are two curves, need to know which to extrapolate for aluminum.
From pg 372 we see that Al has a face centered cubic structure, so extrapolation from copper
seems in order.  The Debye temperature of Cu is 343K, its density is 8.9 kg/m^3
Use constant = (Td(Cu)/d(Cu))*sqrt(M(Cu)/Tm(Cu))=(Td(Al)/d(Al))*sqrt(M(Al)/Tm(Al))
Then use d(Al)=(A/density)^(1/3) to express everything in terms of mass number and density.
Mass numbers for Al = 27 and Cu= 63.6.
Melting temperature for Cu = 1348K.
Td(Al) = Td(Cu)*(d(Al)/d(Cu))*sqrt(A(Cu)*Tm(Al)/(Tm(Cu)*A(Al))
Td(Al)= 343K*sqrt(63.6*934/(27*1348))/(27*8.9/(63.6*2.7))^(1/3)
Td(Al)=  391.76 K
The actual value is 428 K.

#14. Given that the mean free path of electrons in copper at 4K is about 3 mm.
Calculate the resistivity of copper.
From eq 14.68 on pg 385, the resistivity is given by rho= m*vF/(n*e^2*d)
From example 14-7 we see that the  Fermi speed for Cu is 1.6E6 m/s
From example 14-8 the electron density of copper is 8.5E28/m^3
Then rho=9.11E-31 kg*1.6E6 m/s/(8.5E28/m^3*(1.6E-19 C)^2*3E-3)=    2.23284E-13 Ohm m

#29  Density of conductions electrons in silver and B field for Hall effect
Given density 1.05E4 kg/m^3, A=108. Silver has one conduction electron per atom.
a. Density of electrons n=rho*NA/(A*1E-3 kg)=1.05E4 kg/m^3*6.02E23/(108*1E-3 kg) =     5.85278E+28  /m^3
b. Magnetic field for a 1 microvolt Hall voltage if J=1E6 Amp/m^2 flows in sheet of width y=1E-2 m.
The Hall voltage is given by VH= JBy/ne
The required magnetic field is then B= neVH/Jy = (5.85E28 /m^3*1.6E-19 C*1E-6 V)/(1E6 Amp/m^2*1E-2 m)
Required magnetic field =  0.936 Tesla.