How does quantization solve the ultraviolet catastrophe? I understand how classical physics leads to the UV catastrophe. But I cannot understand how quantization solves it. 


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*How can quantization prevent the body from radiating a lot of energy?


I know this question is similar to many other questions but I need a layman's explanation (without too much mathematical details). 


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*Finally why does the peak reach zero intensity after a certain frequency?



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 A: I think you misunderstand the ultraviolet catastrophe. It does not mean that the energy radiated reaches zero at any finite frequency, just that the power tends to zero as the frequency tends to infinity.
Pre-quantum physics thought that blackbody radiation was ruled by the Rayleigh-Jeans law
$$ B_\nu(T) = \frac{2\nu^2 k T}{c^2} $$
This does obviously not go to zero as $\nu \to \infty$.
Now, Planck's law says
$$ B_\nu(T) = \frac{2h\nu^3}{c^2}(\mathrm{e}^{\frac{h\nu}{kT} }-1)^{-1}$$
This goes to zero as $\nu \to \infty$ since exponentials grow faster than polynomials, and it reduces to the Rayleigh-Jeans law in the regime where $\nu$ is "small" such that $(\mathrm{e}^{\frac{h\nu}{kT} }-1)^{-1} \approx \frac{kT}{h\nu}$ holds by Taylor expansion to first order. UV catastrophe averted while reproducing the classical law in the proper limit.
A: Avoiding mathematical formulae to the maximum, and warning for furious hand-waving ahead, I would state it like this:
In the classical picture, there is no quanta concept, so you could have just a little bit of radiation energy at any frequency. However, since quantization appeared, the minimum amount of radiation energy that you could possibly have at a frequency $\nu$ is $h\nu$. Since the temperature regulates the distribution of energy by the Boltzmann equation ($\mathbb{P}(E)\propto e^{-E/kT}$), if you have radiation at a certain frequency such that $h\nu\gg kT$, then having just one photon is "too much", or it is highly unlikely. 
Alternatively, we know from statistical mechanics that it is very unlikely to have radiation with energy higher than $kT$, with the precise expression for this statement given by Boltzmann law. But for $h\nu\gg kT$ type of radiation, Planck leaves us no choice: you have to have more than one photon,  or none. Therefore, you almost do not have this type of radiation (or these type of photons) when the temperature is low, therefore "depopulating" the UV states.
Now I realize something: Planck could have "solved" the UV catastrophe by postulating not complete quantization, but only that the minimum amount of energy at a certain frequency is $h\nu$. With this "floor", you do not need to go further and say that the energy has to come on integer multiples of this minimum quantity, just that, if energy of radiation at this frequency exists at all, then it has a minimum $h\nu$.
A: No, the model given by Planck is no different from classical model, at thermal equilibirium both gives constant average energy. So one can’t use quantum model as said above that average energy is variable and decreasing function, so one gets less participation from higher energy.
Also, if it’s assume the way these models are using now, still it is incorrect to say that quantum model distribute less energy to higher energy level, while it is not. It gives equal energy hνhν to each level but there are lesser modes participance. Or say there are more number of modes at higher level, so energy per mode at higher level becomes less.
The average energy by classical model is,
$⟨ϵ_{classical}⟩=kT$
The average energy by quantum model is,
$\left<ϵ_{quantum}\right>=\dfrac{hν_0}{\exp\frac{hν_0}{kT}−1}$
Both are variable of temperature $T$, and becomes constant at thermal equilibirium. So quantum model doesn’t save UV catastrophe in this form because $ν_0$ is constant. Also frequency of oscillators in a cavity is different and dimension of cavity is constant, that means speed of light is not constant because frequency of oscillators is translated to frequency of emission.
