Why does Planck's law for black body radiation have that bell-like shape? I'm trying to understand Planck's law for the black body radiation, and it states that a black body at a certain temperature will have a maximum intensity for the emission at a certain wavelength, and the intensity will drop steeply for shorter wavelengths. Contrarily, the classic theory expected an exponential increase.
I'm trying to understand the reason behind that law, and I guess it might have to do with the vibration of the atoms of the black body and the energy that they can emit in the form of photons.
Could you explain in qualitative terms what's the reason?
 A: Joshua has beaten me to an answer, but I'll still post this since it's written at a simpler level.
The reason you get a maximum because there are two effects that oppose each other. The number of modes per unit frequency rises as frequency squared, so as long as the energy of the modes is well below kT the energy is proportional to frequency squared. This is why the black body spectrum initially rises approximately as frequency squared.
However the probability that a mode is excited falls exponentially as soon as the energy of the mode is greater than kT, so as the frequency goes to infinity the emitted radiation falls to zero.
The net result of the two effects is that the emission first rises then falls again, and that's why there is a maximum in the middle.
A: The Planck distribution has a more general interpretation: It gives the statistical distribution of non-conserved bosons (e.g. photons, phonons, etc.). I.e., it is the Bose-Einstein distribution without a chemical potential.
With this in mind, note that, in general, in thermal equilibrium without particle-number conservation, the number of particles $n(E)$ occupying states with energy $E$ is proportional to a Boltzmann factor. To be precise:
$$
n(E) = \frac{g(E) e^{-\beta E}}{Z}
$$
Here $g(E)$ is the number of states with energy $E$, $\beta = \frac{1}{kT}$ where $k$ is the Boltzmann constant, and $Z$ is the partition function (i.e. a normalization factor).
The classical result for $n(E)$ or equivalently $n(\lambda)$ diverges despite the exponential decrease of the Boltzmann factor because $g(E)$ grows unrealistically when the quantization of energy levels is not accounted for. This is the so-called ultraviolet catastrophe.
When the energy of e.g. photons is assumed to be quantized so that $E = h\nu$ the degeneracy $g(E)$ does not outstrip the Boltzmann factor $e^{-\beta E}$ and $n(E) \longrightarrow 0$ as $E \longrightarrow \infty$, as it should. This result is of course due to Planck, hence the name of the distribution. It is straightforward to work this out explicitly for photons in a closed box or with periodic boundary conditions (e.g. see Thermal Physics by Kittel).
I hope this was not too technical. To summarize, the fundamental problem in the classical theory is that the number of accessible states at high energies (short wavelengths) is unrealistically large because the energy levels of a "classical photon" are not quantized. Without this quantization, the divergence of $n(E)$ (equivalently, of $n(\lambda)$) would imply that the energy density of a box of photons is infinite at thermal equilibrium. This is of course nonsensical.
A: Given that Plank is a theoretical Physicist, his job was to to use mathematics, known Physics, and one or two assumptions to derive new equations. So, what plank did was the following, he fist defined unambiguously the system he was going to study and that was the blackbody. he knew that the entropy was the key solution, since he knew how to relate the entropy to the energy density and finally to the spectral emittance. So, he postulated that the energy of his system was to be given by U=Pe, where P is in general a large integer and e is chuck of energy, as he wanted to calculate the entropy in Boltzmann's way of doing it. He did that and succeeded in deriving an equation whose results matched well with the experimental measurements.
