Silly Questions said:
"Planck's solution to the ultraviolet catastrophe was not to quantize radiation itself ..."
Whoa. Question answered. Holy smokes. Now it all makes sense.
Of course that brings a new question: is it possible to broadcast a "Planck-invalid" EM wave? Is there any way to emit an EM wave that is impossible for atoms to radiate?
According to the known physics there's neither a longest nor a shortest possible wave length for electromagnetic waves. The wave length (and thus momentum and energy) of electromagnetic waves are continuous also in the quantum theory (quantum electrodynamics).
The solution to the ultraviolet catastrophe does not need a quantization of the energy of the em. field but the quantization of the possible amounts of energy being exchanged with matter (i.e., charged particles) for each wave mode of definite frequency.
It's much easier to understand in the modern quantum-field theoretical formulation than via the very tricky way Planck used to derive his famous formula.
First start with a finite volume. For simplicity we use a cube of length ##L## and assume periodic boundary conditions, i.e., the vector potential in the radiation gauge (which uniquely describes free em. fields) should obey the conditions ##\vec{A}(t,\vec{x}+L \vec{n})=\vec{E}(t,\vec{x})## . Then all you need to know about the quantized electromagnetic field is that it is equivalent to an infinite set of harmonic oscillators.
Each harmonic oscillator describes a field mode with definite momentum ##\vec{p}=\hbar \vec{k}## and definite helicity ##\pm 1## (i.e., right- and left-circular polarized fields). Such a field mode or "photon" has also definite energy ##E(\vec{p})=\hbar \omega=\hbar |\vec{k}|c=|\vec{p}| c##. In our periodic-box setup the momenta are indeed "quantized", i.e., allowed are the wave vectors ##\vec{k}=\frac{2 \pi}{L} \vec{n}## with ##\vec{n} \in \mathbb{Z}##. For each allowed ##\omega(\vec{k})## the possible energies of the corresponding harmonic oscillator are ##E_j=j \hbar \omega## with ##j \in \{0,1,2,\ldots \}##.
Thus the partition sum for the field modes with definite frequency in the cavity with walls kept at given absolute temperature ##T## in thermal equilibrium is
$$Z(\omega,t)=\sum_{n=0}^{\infty} \exp(-\beta j \hbar \omega)=\frac{1}{1-\exp(-\beta \hbar \omega)},$$
where ##\beta=1/(k T)##.
The mean energy in this field mode is
$$\langle E(\vec{k}) \rangle = -\frac{1}{Z} \partial_{\beta} Z=\frac{\hbar \omega}{\exp(\beta \hbar \omega)-1}.$$
Now we take the "thermodynamic limit", i.e., we make the volume very large. Then in any volume of momentum space ##\mathrm{d}^3 \vec{p}## we have ##\frac{2 \mathrm{d}^3 \vec{p} L^3}{(2 \pi \hbar)^3}## field modes (the factor of 2 takes account of the two polarization states). Thus the spectral distribution of the energy is given by
$$\mathrm{d} U=\frac{2 \mathrm{d}^3 \vec{p} L^3}{(2 \pi \hbar)^3} \langle E(\vec{k}) \rangle = \frac{2 \mathrm{d}^3 \vec{k} L^3}{(2 \pi)^3} \frac{\hbar \omega}{\exp(\beta \hbar \omega)-1}.$$
This tells us that since each photon with frequency ##\omega## carries an energy ##\hbar \omega## that photons are massless bosons with the corresponding Bose-Einstein distribution.
A more common form of Planck's Law is to refer it to the energy density per frequency rather than wave-vector interval. Since ##\omega^2=c^2 k^2## we have
$$\omega \mathrm{d} \omega=c^2 k \mathrm{d} k\; \Rightarrow \; \mathrm{d} k=\frac{\mathrm{d} \omega}{c}$$
and thus
$$\mathrm{d}^3 \vec{k}=4 \pi k^2 \mathrm{d} k =\frac{4 \pi}{c^3} \omega^2 \mathrm{d} \omega$$
and thus finally Plancks radiation formula,
$$\frac{\mathrm{d} U}{\mathrm{d} \omega}=\frac{\hbar \omega^3 V}{\pi^2 c^3} \frac{1}{\exp(\beta \hbar \omega)-1}.$$
