Rohlf Chapter 15,#5,6,9,12,17,21,22,24

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
Magnetic permeability of space  mu0 4*Pi()*1E-7 N/A^2 exact
1.25664E-06
Flux quantum  fluxon 2.07E-15 tesla m^2

#5 Estimate critical field necessary to destroy superconductivity in vanadium at 4.2K.
Critical field at zero K is 0.14 T and the critical temperature is 5.4K.
The critical magnetic field follows the relationship Bc=Bc(0)*(1-(T/Tc)^2).
So for 4.2K the critical field is Bc=0.14T*(1-(4.2K/5.4K)^2)=    0.0553 Tesla

#6 Estimate condensation energy of lead cylinder and field required to reduce it by 10%
Critical temperature is 7.2K and critical magnetic field at zero K is 0.08 T.
Lead cylinder of volume 1E-9 m^3 cooled to 4.2K.
a. The critical magnetic field is Bc=0.08T*(1-(4.2K/7.2K)^2)=    0.0528 Tesla
The condensation energy is Bc^2*V/(2*mu0)=(.0528^2*1E-9m^3)/(2*4*Pi()*1E-7)
Condensation energy=  1.10925E-06 Joules
b. The condensation energy is proportional to the square of the magnetic field, so to
reduce it to 90% would require B=sqrt(.9)*Bc=   0.0501 Tesla

#9. Calculate maximum current for a superconducting lead wire.
Wire of diameter 100 micrometers cooled to 4.2K.
Critical temperature is 7.2K and critical magnetic field at zero K is 0.08 T.
This requires the use of the expression for the magnetic field of a long wire,
which is B=mu0*I/(2*pi*r)
First the critical magnetic field at 4.2K must be calculated.
Bc=0.08*(1-(4.2/7.2)^2)=  0.0528 Tesla
Then the critical current is given by I=2*pi*r*Bc/mu0=2*pi*.5E-4m*0.0528/(4*pi*1E-7)
Critical current =   13.194 amperes

#12 Calculate the density of superconducting electrons in tin at zero kelvin.
The magnetic field penetration depth in tin at 0K is given as 30 nm.
The density of superconducting electrons is given by n=mc^2/(4*pi*ke^2*penetration depth^2)
n=511000 eV/(4*Pi()*1.44 eV nm*30^2nm^2)=   31.38   /nm^3 = 3.13766E+28  /m^3

#17. Determine the energy gap for vanadium from the heat capacity data of Figure 15-22.
Assuming heat capacity of the form C=A*exp(b/kT) according to equation 15.46, two points
were chosen and the resulting heat capacity ration solved for b, yielding b=7.4k.
If b represents the change in energy of the electrons as they begin to participate in the
heat capacity, then that energy is approximately half the band gap, presuming that the
effective Fermi level is in the middle of the gap.
With that assumption, then the gap is given by Eg/2= 7.4k
Eg=2*7.4*8.62E-5=  0.0013 eV

#21. Calculate frequency of radiation to cause tunneling current to occur at voltages of 1.35 mV.
Given Pb-I-Pb Josephson junction operated at liquid He temp and with energy gap 2.7 meV.
The frequency of oscillation for a Josephson junction is given by f=2eV/h
f=2*1.6E-19C*1.35E-3 V/4.14E-15 eV s=   6.52E+11  /s

#22. Calculate required area of squid magnetometer.
Given that a fractional change of 1E-4 in Earth's field corresponds to one flux quantum.
The Earth's magnetic field is given at 5x10^-5 T.
The magnetic field change is then 5x10^-9 T and you want to choose the area so that this change
corresponds to one flux quantum, which is 2.07E-15 T m^2.
The area is then A=(2.07E-15 T m^2)/(5E-9 T) =   0.000000414 m^2
4.14E-07 m^2

#24. Calculate voltage across Josephson junction if a frequency of 0.5 GHz is measured.
V=f*h/(2*e)= (0.5E9/s)*(4.14E-15 eV s)/2=   0.000001035 V
1.035E-06 V