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