= Solution
As the <temperature> falls, electrons and protons form neutral <hydrogen> through <cosmological recombination>. The free <electron> fraction drops sharply, and so does the scattering rate $\Gamma_\gamma=n_e\sigma_Tc$. When this rate becomes comparable to or smaller than the <Hubble parameter>, the plasma becomes transparent and <photon decoupling> occurs. The <Thomson scattering> cross-section itself does not abruptly vanish. The rapid change is in the abundance of free targets.
The <Hydrogen binding energy> is $13.6\,\mathrm{eV}$, but recombination occurs at a much smaller thermal energy, of order $0.3\,\mathrm{eV}$, or a few thousand kelvin. There are roughly a billion photons per baryon, so the high-energy tail can keep hydrogen ionized long after $k_BT$ drops below the binding energy. The falling exponential ionizing tail then makes the transition comparatively sharp. Last scattering follows when sufficiently few electrons remain free.<Cosmology by David Tong>[Section 2, recombination and photon decoupling: small baryon-to-photon ratio, thermal tail, and the distinction between recombination and last scattering]{ref}
To estimate a horizon, specify an expansion model as well as the redshift. The proper <particle horizon> at emission is
$$
d_{\mathrm{ph}}(t_*)=a(t_*)\int_0^{t_*}\frac{c\,dt}{a(t)}.
$$
As a transparent analytic estimate, approximate a spatially flat <matter-dominated universe> by $a(t)=(t/t_{\mathrm{now}})^{2/3}$ throughout. Then
$$
d_{\mathrm{ph}}(t_*)=3ct_*,
\qquad
t_*=\frac{t_{\mathrm{now}}}{(1+z_*)^{3/2}}.
$$
Taking the illustrative present value $H_0=70\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$, this model has $t_{\mathrm{now}}=2/(3H_0)\simeq9.3\times10^9\,\mathrm{yr}$. At $z_*=10^3$,
$$
\boxed{
t_*\simeq3\times10^5\,\mathrm{yr},\qquad
d_{\mathrm{ph}}(t_*)\simeq0.27\,\mathrm{Mpc}.
}
$$
This is a proper horizon radius at emission; its corresponding present comoving length is about $270\,\mathrm{Mpc}$.
For the same model the comoving distance to the <Cosmic microwave background last-scattering surface> and its <angular diameter distance> are
$$
\chi_*=3ct_{\mathrm{now}}
\left[1-(1+z_*)^{-1/2}\right],
\qquad
D_A=\frac{\chi_*}{1+z_*}.
$$
Therefore the angular extent associated with one radial horizon length is
$$
\boxed{
\theta_{\mathrm{ph}}\simeq\frac{d_{\mathrm{ph}}}{D_A}
=\frac1{\sqrt{1+z_*}-1}
\simeq0.033\,\mathrm{rad}\simeq1.9^\circ.
}
$$
This is an order-one-degree estimate. Including the earlier <radiation domination> and late <dark energy> changes the coefficient; it does not make the horizon span the sky. A horizon patch diameter has twice this angle, whereas the acoustic scale uses a <sound horizon> and differs again. The redshift by itself, without an expansion model or normalization, is insufficient for a unique horizon distance.
The <horizon problem> is that the observed <cosmic microwave background> has nearly the same mean <temperature> across angular separations vastly greater than this scale. In the purely decelerating <Hot Big Bang model>, sufficiently separated last-scattering regions had neither time to exchange signals nor a common post-Big-Bang causal region that could establish their observed thermal uniformity. The puzzle concerns the missing causal explanation of those initial correlations; it is not an inconsistency of specifying uniform initial conditions mathematically. An earlier period of <cosmic inflation> can enlarge a formerly causally connected region beyond the later observable horizon.
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