Question

The central temperature of white dwarf can be estimated using the relation between temperature gradient and radiative flux. That turns out to be about 5x107 K. Now combined with the radius you calculated in the previous question, evaluate if the degeneracy condition is reached for the electrons and ions in Sirius B. As in the previous question, assume the white dwarf is made of fully ionized carbon.

Answer 1

To evaluate if the degeneracy condition is reached for the electrons and ions in Sirius B, we need to compare the central temperature of the white dwarf with the critical temperature for degeneracy.

Step-by-step explanation:

To evaluate if the degeneracy condition is reached for the electrons and ions in Sirius B, we need to compare the central temperature of the white dwarf with the critical temperature for degeneracy. The critical temperature for electrons is given by:

kTc = (h^2 / 2πme) (3/(8π))^(2/3) n^(2/3)

Where k is the Boltzmann constant, h is the Planck constant, me is the mass of the electron, and n is the number density of the electrons. For fully ionized carbon, the number density of electrons is equal to the number density of carbon nuclei. Now, let's substitute the values:

k = 1.38 × 10^(-23) J/K

h = 6.626 × 10^(-34) J·s

me = 9.11 × 10^(-31) kg

n = (ρ / (12 u)) (6.022 × 10^23 particles/mol)

Where ρ is the density of fully ionized carbon and u is the atomic mass unit. Let's assume the density of fully ionized carbon is equal to the density of solid carbon, which is approximately 2.267 g/cm³. Substituting these values, we can calculate the critical temperature for degeneracy.

If the central temperature of the white dwarf is higher than the critical temperature, then the degeneracy condition is reached for the electrons and ions.