When a star has nearly burned up its internal fuel, it may become a white dwarf. It is cru...

When a star has nearly burned up its internal fuel, it may become a white dwarf. It is crushed under its own enormous gravitational forces to the point at which the exclusion principle for the electrons becomes a factor. A smaller size would decrease the gravitational potential energy, but assuming the electrons to be packed into the lowest energy states consistent with the exclusion principle, "squeezing" the potential well necessarily increases the energies of all the electrons (by shortening their wavelengths). If gravitation and the electron exclusion principle are the only factors, there is a minimum total energy and corresponding equilibrium radius.

(a) Treat the electrons in a white dwarf as a quantum gas. The minimum energy allowed by the exclusion principle (see Exercise 67) is

Note that as the volume V is decreased, the energy does increase. For a neutral star, the number of electrons, N, equals the number of protons. Assuming that protons account for half of the white dwarf's mass M (neutrons accounting for the other half), show that the minimum electron energy may be written

where R is the star's radius

(b) The gravitational potential energy of a sphere of mass M and radius R is given by

Taking both factors into account, show that the minimum total energy occurs when

(c) Evaluate this radius for a star whose mass is equal to that of our Sun, ~2 x 1030 kg.

(d) White dwarfs are comparable to the size of Earth. Does the value in part (c) agree?