How does Planck's quantum theory explain the low intensity of radiation for high frequencies?

I.e why does the black body curve become lower at the high frequency side?

• have a look : it depends on the input hypothesis ( continuous versus quantized) hyperphysics.phy-astr.gsu.edu/hbase/mod6.html diferent functional forms appear – anna v Sep 28 at 5:37
• Please explain it in word not mathematically – Debakant Sep 28 at 5:43
• It all depends on the mathematics, when there is not quantization the mathematics shows the ultraviolet effect, when there is the values are limited by the function. The data follwo the quantized curve, so it means the mathematics describes it. I means that the higher the frequency in classical mechanics the more energy one can fit in the form of electromagnetic waves, whereas the data shows the opposite, that there is a peak and then a fall in the energy, agreeing with quantized values.. the link states it in words also. – anna v Sep 28 at 8:54
• Thanks But as the result is only mathematically so how scientists at that time accept that – Debakant Sep 28 at 9:54
• The ultraviolet catastrophy was not accepted as physical because the data showed a different behavior. Thats why the black body was accepted, because it fitted the data. – anna v Sep 28 at 10:44

Since you asked for a non-mathematical explanation why Planck's blackbody spectrum becomes lower at the high-frequency side, I will give it a try with as little mathematics as possible.

The free space can have electromagnetic waves with all directions, and with all frequencies (from $$0$$ to $$\infty$$).

Planck's important postulate is that each electromagnetic oscillation with frequency $$\nu$$ cannot have any arbitrary energy $$E$$, but only integer multiples of $$h\nu$$. $$E=nh\nu, \quad \text{with } n=0,1,2,3,\dots \tag{1}$$

Let us consider some example frequencies: $$100$$ THz, $$200$$ THz, $$300$$ THz, ... , $$900$$ THz.

Then the energy quanta $$h\nu$$ of these frequencies are: $$0.41$$ eV, $$0.82$$ eV, $$1.23$$ eV, ... , $$3.7$$ eV.

So we get for each frequency a ladder of energy levels. The lower frequencies having a dense energy spacing, and the higher frequencies having a coarse energy spacing.

Now, let this ensemble of oscillation modes be heated to a temperature $$T$$. This will cause the oscillators to get excited with energies between $$0$$ and roughly a few $$kT$$.

Many oscillators have energy $$E=0$$, less oscillators have energy $$E=h\nu$$, even less have energy $$E=2h\nu$$, etc. The exact percentages can be calculated by Boltzmann's distribution, but that would need too much mathematics here. Therefore I visualize the percentages for an example temperature $$T=14000$$ K ($$\Rightarrow$$ with thermal energy $$kT=1.2$$ eV) in the image below (by 10 blue dots for each frequency).

You see the oscillators are excited with energies between $$0$$ and roughly $$3$$ eV ($$= 2.5\ kT$$).

Looking carefully you see:

• For low frequencies (the left part, where the energy spacing $$h\nu$$ is smaller than the thermal energy $$kT$$) the average energy $$\overline{E}(\nu,T)$$ is nearly independent of the frequency $$\nu$$. It is approximately $$kT$$.
• For high frequencies (the right part, where the energy spacing $$h\nu$$ is larger than the thermal energy $$kT$$) the average energy $$\overline{E}(\nu,T)$$ becomes smaller and then approaches zero. This is simply because for high frequencies there are no energy levels available immediately above zero.

The reasoning above may be intuitive (hopefully), but it is not very precise. Fortunately the same reasoning can be done in a more mathematical way (see for example "The Derivation of the Planck Formula", page 9-10). Then the result for the average energy is: $$\overline{E}(\nu,T)=\frac{h\nu}{e^{h\nu/kT}-1} \tag{2}$$ From this formula (2) you can get as approximations for small and large frequencies again the same features, which we could extract from the image above in a hand-waving manner.

You can recognize this average energy (2) as part of Planck's law for the black-body radiation $$B_\nu(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e^{h\nu/kT}-1}. \tag{3}$$ The other factor $$\frac{2\nu^2}{c^2}$$ can be accounted for the number of oscillator modes per frequency range and per volume (see for example "The Derivation of the Planck Formula", page 3-5).

Planck's theoretical postulate (1) of quantized oscillator energies surely seems bold at first glance. But at the end it leads to a spectral curve (3) for the black-body radiation which is in excellent agreement with experimental measurements. And this is the ultimate justification for the theory.