Past exam of the radiative processes and high energy astrophysics course of the University of Oxford / 2024 / 2 / b / Solution

Charge neutrality in fully ionized hydrogen gives $n_e=n_p$ and $\rho =(m_p+m_e)n_e\simeq m_p{n}_e$. With only Thomson scattering, the photon mean free path is $\ell =(n_e{\sigma }_T{)}^{-1}$. HenceHere and below the tiny electron mass correction can be retained by replacing $m_p$ by $m_p+m_e$. $L$ is the central mean free path, not a luminosity. The divergence at $r=R$ reflects the vanishing electron number density.

Using $M_{\odot }=1.9885\times 10^{30}\mathrm{kg}$, $R_{\odot }=6.957\times 10^8\mathrm{m}$, $m_p=1.6726\times 10^{-27}\mathrm{kg}$, and ${\sigma }_T=6.6525\times 10^{-29}{\mathrm{m}}^2$ givesThese are this profile's values, rather than the Sun's actual central mass density or opacity profile.