= Solution
Use spectral <radiant exitance> per unit wavelength. The <Planck law> is
$$
M_\lambda(T)=\frac{2\pi hc^2}{\lambda^5}\frac1{e^{hc/(\lambda k_{\rm B}T)}-1}.
$$
A hotter <blackbody> radiates more at every wavelength, its wavelength peak moves to shorter wavelengths according to <Wien's displacement law>, and its integrated flux rises as $\sigma T^4$ by the <Stefan–Boltzmann law>. The wavelength convention matters: a spectrum plotted per unit frequency has a different peak transformation.
\Image[/past-exam-of-the-atmospheric-physics-course-of-the-university-of-cambridge/2013/paper-1-ap2013-blackbody-spectra.png]
{title=Illustrative wavelength spectra: a hotter blackbody has a larger total flux and a shorter peak wavelength}
{source=paper-1-ap2013-blackbody-spectra.py}
For <planetary equilibrium temperature>, assume an isotropic blackbody star, a circular orbit, negligible internal heating, unit surface thermal emissivity and efficient global heat redistribution so the surface has one temperature. The received stellar irradiance is $F_*=\sigma T_*^4(R_*/a)^2$. A planet of radius $R_p$ intercepts $\pi R_p^2F_*$ but radiates over $4\pi R_p^2$, giving the <projected-area factor in planetary energy balance>
$$
\boxed{T_0=T_*\sqrt{\frac{R_*}{2a}}=303.4\,\mathrm K.}
$$
Without the uniform-temperature assumption, this calculation gives $\langle T^4\rangle^{1/4}$, not a unique arithmetic mean of surface temperature.
With <Bond albedo> $A=0.05$, the absorbed fraction is $1-A$. Hence
$$
\boxed{T_A=T_0(1-A)^{1/4}=299.5\,\mathrm K,}
$$
a decrease of approximately $3.87\,\mathrm K$. This albedo weights incident stellar light and does not imply thermal emissivity $0.95$.
Define the <planetary effective emissivity> by outgoing top-of-atmosphere flux $F_{\rm out}=\epsilon_{\rm eff}\sigma T_s^4$. Maintaining the same absorbed stellar power with $T_s=312\,\mathrm K$ gives
$$
\boxed{\epsilon_{\rm eff}=\left(\frac{T_A}{T_s}\right)^4=0.849.}
$$
The <greenhouse effect> allows thermal radiation to escape from colder layers, so the outgoing flux is less than a black surface at $T_s$ would emit. Downward atmospheric emission contributes to the surface energy balance; this does not violate the top-of-atmosphere energy balance.<Climate and Earth's Energy Budget by NASA Earth Observatory>[Thermal radiation, atmospheric absorption and effective radiating versus surface temperature]{ref}
A <single-layer greenhouse model> makes the distinction quantitative. Take a shortwave-transparent, isothermal layer with infrared absorptivity and emissivity $\epsilon_a$, above a surface black in the thermal infrared. Writing $Q=(1-A)F_*/4$, its surface and atmospheric balances are
$$
Q+\epsilon_a\sigma T_a^4=\sigma T_s^4,\qquad
\epsilon_a\sigma T_s^4=2\epsilon_a\sigma T_a^4.
$$
For $\epsilon_a>0$, $T_a^4=T_s^4/2$. Radiation escaping to space is $(1-\epsilon_a)\sigma T_s^4+\epsilon_a\sigma T_a^4$, so
$$
\boxed{\epsilon_{\rm eff}=1-\frac{\epsilon_a}{2},\qquad
\epsilon_a=0.301,\qquad T_a=262.4\,\mathrm K.}
$$
Thus \b[$0.849$ is the planet's effective emissivity; $0.301$ is the layer's infrared emissivity within this particular model.] They are not interchangeable. Real atmospheres require wavelength-dependent absorption, a vertical temperature profile and nonradiative heat transport; the measured temperatures alone do not uniquely retrieve an atmospheric optical profile.
\Image[/past-exam-of-the-atmospheric-physics-course-of-the-university-of-cambridge/2013/paper-1-ap2013-greenhouse-fluxes.png]
{title=One-layer radiative equilibrium distinguishes planetary effective emissivity from the infrared emissivity of the atmosphere}
{source=paper-1-ap2013-greenhouse-fluxes.py}
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