--class 15--
to get the formula for "degeneracy pressure" wiki it!

Find the mass density of Cu and the mass of a Cu atom.
Use this to calculate the number of Cu atoms per m^3.
Assume there are 2 valence electrons from the Cu that are "free".
Calculate the electron number density, and the corresonding "degeneracy pressure".
What is the result in atm?

wiki "Fermi gas" and find (about half way down) the relation between the
Fermi energy: E_F and the number of particles per volume, N/V
If the Fermi energy is 0.1 mc^2, we must be getting close to the relativisitic case.
For an electron gas find the N/V that is this limit.  Comapre it to the Cu number density

P = (3 pi^2)^(2/3)  hbar^2/(5 m) (N/V)^(5/3)

E_F = hbar^2/(2 m) (3 pi^2 N/V)^(2/3)

Cu density: 8960 Kg/m^3
atomic weight: 63.546 amu

? 8960/(63.546*amu)*2
    .1698246375620825E+30 			!(# electrons/m^3)
? (3*pi^2)^(2/3)*hbar^2/(5*me)*res^(5/3)
    121705015288.5548    			!N/m^2 FYI: about the same as Cu Youngs Modulus
? res/atm
    1201135.112643027    			!atm

if .1 mc^2= E_F = hbar^2/(2 m) (3 pi^2 N/V)^(2/3)
(.2 m^2 c^2)/hbar^2 = (3 pi^2 N/V)^(2/3)
((.2 m^2 c^2)/hbar^2)^3/2/(3 pi^2)=(N/V)

? (.2*me^2*c^2/hbar^2)^(3/2)/(3*pi^2)
    .5245955559691843E+35			!(# electrons/m^3)
? res/.1698246375620825E+30
    308904.2694275787				! multiple of Cu electron density