Black body radiation curve See images about black-body radiation.
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?
 A: 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
$$B_\nu(\nu,T)=\frac{2\nu^2}{c^2}\frac{h\nu}{e^{h\nu/kT}-1}. \tag{3}$$
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.
