= Solution
The <internal effective temperature of a planet> $T_{\rm int}$ parametrizes intrinsic cooling, while the <irradiation temperature> $T_{\rm irr}$ parametrizes incident stellar flux before the redistribution factor $f$. The quantity
$$
\gamma=\frac{\kappa_{\rm vis}}{\kappa_{\rm IR}}
$$
is the visible-to-thermal mean-opacity ratio, and $\tau$ is downward thermal <optical depth>. Under the Eddington two-stream boundary condition,
$$
a=\frac34,
\qquad b=\frac23,
\qquad\boxed{ab=\frac12},
$$
while the common semi-grey choice is $c=\sqrt3$.
For a young giant at $40\,\mathrm{au}$, stellar heating is weak and a still-large $T_{\rm int}$ dominates. Its pressure-temperature profile rises steadily inward, approximately as $T^4\propto b+\tau$, and joins a deep convective adiabat.
For a hot Jupiter close to a Sun-like star, $T_{\rm irr}\gg T_{\rm int}$. It has a broad, nearly isothermal irradiated radiative layer, followed by a deep rise where intrinsic flux and increasing opacity matter. If $\gamma>1$, absorption of starlight above the thermal photosphere can create an <atmospheric thermal inversion>.
For a temperate sub-Neptune around an M dwarf, irradiation and internal cooling can be more comparable. Its profile generally has a moderate radiative layer above a convective interior; near-infrared stellar radiation, molecular opacity, clouds, and hazes determine whether the upper profile is weakly inverted or decreases outward.
Solved by gpt-5.6-sol high.
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