= Solution
In <local thermodynamic equilibrium>, thermal intensity samples the <Planck function> near an optical depth of order unity. The <Eddington-Barbier relation> makes this explicit: $I_\lambda(0,\mu)\simeq B_\lambda[T(\tau_\lambda=\mu)]$. A molecular band has greater <opacity> than its adjacent continuum and therefore samples a higher layer. A band in emission relative to the continuum implies that this higher layer is hotter: \b[the line-forming region has an <atmospheric thermal inversion>] under the assumed LTE, thermal interpretation.
The continuum is thermal radiation from an optically thick, deeper <photosphere>, with comparatively smooth <opacity>. In an H/He <hot Jupiter>, <collision-induced absorption> by H2-H2 and H2-He collisions supplies an important continuum; weak overlapping molecular lines and opaque <exoplanet clouds> can contribute too. It is not a separate <blackbody> emitter floating above the gas. A strongly isothermal layer would erase LTE molecular contrast rather than generate emission peaks.
In the emitting inversion, $dT/dz>0$, whereas the <dry adiabatic lapse rate> has $dT/dz=-g/C_p<0$. Therefore
$$
\boxed{\frac{dT}{dz}>0>-\frac g{C_p},}
$$
which lies on the stable side of the <Schwarzschild criterion>. An upward-displaced parcel cools and becomes denser than the ambient hot upper gas. The region can thus carry and redistribute thermal energy by <radiative transfer>, \b[not by unstable thermal <convection>]. Winds may transport energy horizontally; stability rules out the specified buoyant vertical <convection>, not every possible motion.
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