Here is an interesting commercial site that sells PHOTON GENERATORS of almost any desired range of frequencies:
https://www.kmlabs.com/en/wavelength-to-photon-energy-calculator
It seems to be a serious technical company and, on this page, it provides
a calculator that relates properties of photons: energy and wavelength (therefore frequency) plus flux power and flux rate of photons.
The formulae that they use (are displayed) are elementary:
E = h.f = h.c/L (in Ev)
P = F.E (in uW), where F is the flux of photons per second.
You can play with any of the four entries, and the others relate immediately.
But, to obtain the frequency, you have to do the math with f = c/L.
What made me wonder was how come a photon has wavelength, as it's a
transversal measurement of an EM waveform, which brings me to Maxwell,
because I can't imagine a photon with longitudinal (time) and transversal
amplitude (EM fields), which causes to me a dilemma.
IF a photon has such manifestations, then it occupies a site in the space
which varies widely with frequency.
1- You'd have photons 1 meter long at 300 Mhz.
2- You'd have photons 1 cm long at 30 Ghz.
3- You'd have photons 500 nm long at 600 Thz (visible light).
4- You'd have photons 50 nm long at 6 Phz (UV light).
5- You'd have photons 100 pm long at 3 Ehz (X-rays)
and so on. In 4 and 5, the photon can ionize atoms easily, even when the
average atom size is 1 Armnstrong (0.1 nm).
So, how come an electron of an H atom can get the whole h.f energy of a
photon type 4 if the cross-section of such photon is (avg.) 500 times larger
than the accepted average size of an atom (not to mention the electron, which is sized down this value by 100,000,000 times?).
At this paper, the cross-section of EM radiation is defined by the frequency
at which molecules absorb EM radiation for any reason, and relate it to the
wavelength of the EM radiation. It can happens at microwave regions and
far beyond, without ionization required.
So, a long way since simplistic models in old QM, and also shocking
co-existence of the duality wave-particle (real or not).
https://nebula.esa.int/sites/default/files/neb_study/158/C11340ExS.pdf
But, digging into the maxwellian world, for a given EM single ray, it is that:
The effective area of an isotropical antenna is given as
A = 0.08 λ²
Being that E and H fields of an EM wave carry, each one, half the intensity of
a planar EM wave (sinusoidal), the total intensity can be put in terms of E field only. So, for a sinusoidal EM wave of frequency f (
Iavg = c.εo.Eo²/2
where c is the speed of light, εo is the permittivity of free space, and Eo is the maximum electric field strength; intensity, as always, is power per unit area (here in W/m2).
In this case, the power captured by an isotropical antenna is about:
Pavg = 0.04 λ².c.εo.Eo² (in Watts)
But, also, the power conveyed by a single ray of EM radiation (emitted by a
planckian resonator), E=h.f, so
Pem = h.f/τ (E is the energy per wavelength, f its frequency and τ
the duration of the EM wave). Unit is Watt.
then, equating Pavg and Pem, it gives
0.04 λ².c.εo.Eo² = h.f/τ
and, finally
h.f = 0.04 τ.λ².τ.c.εo.Eo²
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I wonder if the unit of energy h.f involved in an absorption or emission of
a "quanta of energy° can be related in this way with the EM properties of
the burst of EM energy, with a duration of τ seconds, providing that λ and f
satisfy c = λ.f
It's understood that quantum emission and absorption take a time in the
region of nanoseconds.
My questions are:
- Can be possible that τ seconds represent the lower value of
emission or absorption of EM energy in the atom? If, for absorption,
τ last more than the actual time involved? The reciprocal question
also applies for the emission phenomenon.
- Can be possible that, for shorter times, the single ray (quanta) use
its excess of energy to give part of it to other constituents of the atom?
Or such energy vanishes in the quantum vacuum.
- At any case, what actually represent the wavelength of a photon?