= Solution
For an isotropic <degenerate electron gas>, pressure is one third of the momentum flux. In the ultrarelativistic limit each occupied mode has speed $c$, so the supplied momentum-space state count gives
$$
n_e=\int_0^{p_0}\frac{8\pi p^2}{h^3}dp=\frac{8\pi p_0^3}{3h^3},
\qquad
P_e=\frac13\int_0^{p_0}pc\frac{8\pi p^2}{h^3}dp
=\frac{2\pi c p_0^4}{3h^3}.
$$
A fully ionized helium nucleus provides two electrons for approximately $4m_p$ of mass, so $n_e=\rho/(2m_p)$. Eliminating the <Fermi momentum> $p_0=(3h^3\rho/(16\pi m_p))^{1/3}$ therefore gives
$$
\boxed{P_e=K\rho^{4/3},\qquad
K=\left(\frac3{2\pi}\right)^{1/3}\frac{hc}{16m_p^{4/3}}.}
$$
The two-spin-state degeneracy is already included in the supplied state count; no further factor 2 is needed.
If <electron degeneracy pressure> supplies the star's pressure, this is an index-three <stellar polytrope>. Its central pressure and density obey
$$
c_3\frac{GM^2}{R^4}
=K\left(\frac{3a_3M}{4\pi R^3}\right)^{4/3}.
$$
The radius cancels, leaving $M^{2/3}=K(3a_3/(4\pi))^{4/3}/(c_3G)$. Thus the <Chandrasekhar mass from central polytropic constants> is
$$
\boxed{M=\left(\frac3{2\pi}\right)^{1/2}
\left(\frac{3a_3}{4\pi}\right)^2
\frac1{64c_3^{3/2}}
\left(\frac{hc}{G}\right)^{3/2}\frac1{m_p^2}.}
$$
The dependence on microscopic constants reflects the balance of fermionic pressure and gravity. The absence of radius dependence is special to the limiting $4/3$ pressure exponent. Ordinary partially relativistic <white dwarfs> can have a range of masses below this <Chandrasekhar limit>; the formula does not force every <white dwarf> to have this mass. As a degenerate stellar core grows toward the limiting mass, compression cannot provide an arbitrarily larger supporting pressure relative to gravity. Above it the ideal cold, nonrotating electron-supported configuration cannot remain in equilibrium; nuclear ignition, collapse or other changes in the support mechanism must intervene. Which outcome occurs depends on the core's composition and evolutionary state.
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