Matter inside a ''white dwarf'' is in a state of degeneracy, and the dependence of the pressure on the density is of the form `P=Arho^(5//3)`, where P is the pressure and `rho` is the density. Find the expression for the constant A and show that the pressure is due to the electron gas, the pressure due to the heavy particles being negligible.

Answer

`A=h^(2)/(20m_(e)(m_(alpha)//2)^(5//3))~~3.2xx10^(6)Pa*m^(5)*kg^(5//3)`

Explanation

The pressure in the ..white dwarf.. is due to gas consisting of free electrons and of helium nuclei. These particles are in a degenerate state, like free electrons in a metal. The mass of an electron is almost 1/8000 the mass of a helium nucleus. Therefore the Fermi energy of the electron and the pressure of the electron gas are 8000 times greater than the corresponding quantities for helium. Hence the helium pressure may be neglected. Making use of the result of the previous problem for the pressure of the electron gas, we obtain
`PV^(5//3)=h^(2)/(5m_(e))(3/(8pi))^(2//3)N_(e)^(5//3)`
There are two electrons to each helium nucleus, so `N_(e)=2N_(alpha)=2M//m_(alpha)` where M is the mass of the star, and `m_(alpha)=4.002xx1.66xx10^(-27)kg` is the mass of a helium nucleus.
Hence `PV^(5//3)=AM^(5//3),orP=Arho^(5//3)`, where
`A=h^(2)/(5m_(e)(m_(alpha)//2)^(5//3))(3/(8pi))^(2//3)=3.2xx10^(5)Pa*m^(5)*kg^(5//3)`