= Solution
The coefficient $j_\nu$ is <power> per unit <volume>, <frequency> and <solid angle>. For uniform, <optically thin>, <isotropic> emission from $V=4\pi R^3/3$, the <spectral luminosity> is $L_\nu=4\pi Vj_\nu$. With <photon> <energy> $\varepsilon=h\nu$, <conservation of energy> and the change of spectral variable give
$$
L_\nu\,d\nu=\varepsilon L_\varepsilon^X\,d\varepsilon,
\qquad d\nu=\frac{d\varepsilon}{h}.
$$
Thus
$$
\boxed{L_\varepsilon^X=\frac{4\pi}{h\varepsilon}\frac{4\pi R^3}{3}
j_0\frac{2}{1+x_H}
\left(\frac{n_e}{10^{-3}\,\mathrm{cm}^{-3}}\right)^2
\sqrt{\frac{T_0}{T}}e^{-\varepsilon/(k_BT)}.}
$$
The $4\pi$ outside $V$ integrates over emitted <solid angle>; $1/\varepsilon$ changes radiated <energy> to a <photon> count, and $1/h$ is the <Jacobian determinant> from <frequency> to <energy>.
With $h$ in $\mathrm{J\,s}$ and $\varepsilon$ in joules, this formula produces $\mathrm{J^{-1}\,s^{-1}}$. To compare with counts per $\mathrm{keV}$, multiply by $1.602176634\times10^{-16}\,\mathrm{J/keV}$. Equivalently, the conversion cancels the same joule-per-$\mathrm{keV}$ factor in $\varepsilon$: one can use the numerical $\varepsilon$ in $\mathrm{keV}$ in the $1/\varepsilon$ factor, retain $h$ in $\mathrm{J\,s}$, and express $k_BT$ in $\mathrm{keV}$ in the exponential. The radius in the <volume> must still be converted to metres because $j_0$ is per $\mathrm{m^3}$.
Back to article page
