= Solution
\b[Buoyancy and the convective criterion.] Consider a small fluid parcel displaced outward by $\delta r$. Assume that sound waves keep its <pressure> equal to the ambient <pressure>, while heat exchange is slow enough that its displacement is an <adiabatic process>. Composition is uniform, so the parcel and its surroundings have the same <mean molecular weight>. For an <ideal gas>, $P\propto\rho T$, and an adiabatic parcel obeys $P\propto\rho^\gamma$. If $\delta\ln P<0$ is the <pressure> change on moving outward, then
$$
\delta\ln\rho_{\rm parcel}=\frac1\gamma\delta\ln P,\qquad
\delta\ln\rho_{\rm ambient}=(1-\nabla)\delta\ln P.
$$
The <density> excess of the parcel at its new location is therefore
$$
\delta\ln\rho_{\rm parcel}-\delta\ln\rho_{\rm ambient}
=\left[\nabla-\frac{\gamma-1}{\gamma}\right]\delta\ln P.
$$
If $\nabla>(\gamma-1)/\gamma$, this excess is negative: the outward-displaced parcel is lighter and accelerates farther outward. A downward displacement similarly continues downward. Conversely, a smaller gradient gives a restoring buoyancy force. The <Schwarzschild criterion> is thus stability for $\nabla<\nabla_{\rm ad}$ and instability for $\nabla>\nabla_{\rm ad}$, where the <adiabatic temperature gradient> is $\nabla_{\rm ad}=(\gamma-1)/\gamma$. For a monatomic gas,
$$
\boxed{\gamma=5/3\quad\Longrightarrow\quad\nabla_{\rm ad}=2/5.}
$$
A composition gradient would require the additional <density> effect in the <Ledoux criterion>; it is absent in this calculation.
\b[<Opacity> and <pressure>.] The supplied <opacity> is characteristic of a cool hydrogen-rich <stellar atmosphere> dominated by <negative hydrogen ion opacity>. Mostly neutral hydrogen can attach electrons supplied by partially ionized metals; bound-free and free-free absorption involving the resulting $H^-$ ions produces a steep positive <temperature> dependence. The weak <density> dependence and strong <temperature> sensitivity can be represented by local power-law fits. At fixed composition the given <pressure> form is equivalent to $\kappa\propto\rho^{1/2}T^{17/2}$, because $P\propto\rho T$. This is an approximate fit over the relevant cool-atmosphere regime, not an exact <opacity> law at all temperatures; molecular <opacity> or hydrogen ionization eventually changes the regime.
For a geometrically thin atmosphere take the <surface gravity of a star> to be constant, $g_s=GM/R^2$. Combining $dP/dr=-\rho g_s$ with $d\tau/dr=-\kappa\rho$ gives <hydrostatic equilibrium in optical depth>,
$$
\frac{dP}{d\tau}=\frac{g_s}{\kappa_0P^{1/2}T^8}.
$$
The <grey atmosphere> relation gives $T^8=T_e^8(3\tau+2)^2/16$. With negligible external <pressure>, $P(0)=0$, integration yields
$$
\begin{aligned}
P^{3/2}(\tau)&=\frac{24g_s}{\kappa_0T_e^8}\int_0^\tau\frac{du}{(3u+2)^2}\\
&=\frac{8GM}{\kappa_0R^2T_e^8}\left(\frac12-\frac1{3\tau+2}\right).
\end{aligned}
$$
An imposed nonzero external <pressure> would add $P(0)^{3/2}$; the boundary assumption is what fixes the integration constant here.
The factor in parentheses is $3\tau/[2(3\tau+2)]$. Hence
$$
\frac{d\ln T}{d\tau}=\frac3{4(3\tau+2)},\qquad
\frac{d\ln P}{d\tau}=\frac4{3\tau(3\tau+2)},\qquad
\nabla_{\rm rad}=\frac{9\tau}{16}.
$$
The <grey-atmosphere convection onset with eighth-power temperature opacity> occurs when this reaches $2/5$, so
$$
\boxed{\tau_c=\frac{32}{45}.}
$$
The radiative continuation below this boundary would be unstable. Efficient <convection> instead keeps the deeper star approximately adiabatic.
\b[Matching to the convective interior.] Write $P=K\rho^{5/3}$. Eliminating <density> using the <ideal gas law> gives
$$
T=\frac\mu{\mathcal R}K^{3/5}P^{2/5},\qquad
K\propto T^{5/3}P^{-2/3}
$$
at fixed composition. Since $\tau_c$ is a fixed number, the matching <temperature> $T_b$ is proportional to $T_e$, while the matching <pressure> satisfies $P_b^{3/2}\propto g_sT_e^{-8}$. Consequently
$$
K\propto T_e^{5/3}\left(g_s^{2/3}T_e^{-16/3}\right)^{-2/3}
=T_e^{47/9}g_s^{-4/9}.
$$
The <polytropic mass-radius relation> for index $3/2$ gives $K\propto RM^{1/3}$. Raising the matching relation to the ninth power and using $g_s\propto M/R^2$ therefore gives
$$
T_e^{47}\propto K^9g_s^4\propto R^9M^3\frac{M^4}{R^8}=RM^7.
$$
Finally the <Stefan–Boltzmann law> $L=4\pi R^2\sigma_{\rm SB}T_e^4$ yields the <Hayashi relation with eighth-power temperature opacity>,
$$
\boxed{L\propto R^{98/47}M^{28/47}.}
$$
Composition and the <opacity> normalization are fixed in this comparison. The <entropy> coefficient $K$ is spatially constant in each convective model, but need not be the same for stars with different masses or radii.
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