Answer to Question #281840 in Mechanical Engineering for john luvis

Question

White dwarf stars are essentially plasmas of free electrons and free protons

(hydrogen atoms that have been stripped of their electrons). Their densities

are typically 5 × 10^9 kg/m3 (i.e., 10^6 times the average density of Earth).

Their further collapse is prevented by the fact that the electrons are highly

degenerate, that is, all the low-lying states are filled and no two identical

electrons can be forced into the same state.

(a) Estimate the temperature of this system. (Assume nonrelativistic electrons and that p2

f /2m = (3/2)kT .)

(b) Now suppose that the gravity is so strong that the electrons are forced

to combine with the protons, forming neutrons (with the release of a

neutrino). If the temperature remains the same, what would be the density

of this degenerate neutron star?

Answer 1

Density of star=$5*10^{9}kg/m^3$

Mass of star, $M=N(m+2m_p)\equiv2Nm_p$

Where; m=mass of electron, m_p= mass of proton,

Electron density, $n=\frac{N}{v}=\frac{\rho}{2m_p}$

=$\frac{5*10^9}{2*1.67*10^{-27}}=1.49*10^{36}$

a) $T=\frac{P^2}{3mk}$

$=(\frac{3n}{8π})^{\frac{2}{3}}\frac{h^2}{3*9.1*10^{-31}*1.38*10^{-23}}$

$=(\frac{3*1.49}{8π})^{\frac{2}{3}}\frac{10^{24}*h^2}{3*9.1*10^{-31}*1.38*10^{-23}}$

$=\frac{0.316*10^{24}*10^{54}}{3*9.1*1.38}*(6.62*10^{-24})^2$

$=3.67*10^9K$

b) $T=\frac{P^2}{3m_nk}$

Where m_n= mass of neutron

$T=(\frac{3n}{8π})^{\frac{2}{3}}\frac{h^2}{3*1.67*10^{-27}*1.38*10^{-23}}$

$=\frac{0.316*10^{24}*10^{54}}{3*1.67*1.38}*(6.62*10^{-24})^2$

$=2*10^6K$