[mxplxxx (OP)]

How do we measure the energy of a photon? In other words how do we validate the equation E=hf?

[yor_on]

You measure the frequency and get to a 'photon energy'. Planck's 'black body' is a experimental proof of light having a wave/particle duality that you can use for measuring a 'photons' energy, extended by Einsteins 'photoelectric effect' .  http://spiff.rit.edu/classes/phys314/lectures/photoe/photoe.html

[chiralSPO]

photoelectric effect is a good way to measure energy of single photons of ionizing radiation (UV and x-ray)

for visible and infrared light, you can use semiconductor devices (essentially photovoltaics) and measure the voltage produced by the frequencies.

[evan_au]

In the photoelectric effect, above a threshold frequency, an electron is ejected from the surface of a metal, with a kinetic energy that is proportional to the frequency of the incident light. This experiment must be conducted in a vacuum.

The slope of this line is h, Planck's constant.

By using a voltage to oppose the velocity of the emitted electron, you can get a direct measure of the electron's kinetic energy in units of electron-volts (eV).

See the graph at: https://en.wikipedia.org/wiki/Photoelectric_effect#Mathematical_description

[mxplxxx (OP)]

You measure the frequency and get to a 'photon energy'. Planck's 'black body' is a experimental proof of light having a wave/particle duality that you can use for measuring a 'photons' energy, extended by Einsteins 'photoelectric effect' .  http://spiff.rit.edu/classes/phys314/lectures/photoe/photoe.html

Thx. So, I am thinking we can't measure a photon's energy directly. So, how do we know that the E in E=hf is valid? Chicken and egg isn't it?

[alancalverd]

There's a very simple school experiment: measure the lowest voltage at which a light-emitting diode produces light. That tells you the energy E of those photons in electron volts. Now use a spectrometer to measure the wavelength λ of that light. E = hf = hc/λ by definition.


For a bit more sophistication you can generate x-rays with a known voltage on the x-ray tube anode, and use a crystal to diffract the x-rays. If the anode is,say, molybdenum, you will get a sudden increase in the x-ray beam intensity as the anode voltage exceeds 17,300V. This Kα radiation will produce a characteristic bright spot on the diffraction pattern, from which you can calculate the wavelength of a 17.3 keV photon.

[mxplxxx (OP)]

E = hf = hc/λ by definition.

Good to know, but it seems to me that none of these arguments involve directly measuring the energy E in joules of a single photon as per the E=hf equation. 

[chiralSPO]

it seems to me that none of these arguments involve directly measuring the energy E in joules of a single photon as per the E=hf equation. 

That is correct. There is no good way to directly measure the energy of a single photon.

But, if you go back and look at the answers that were given to you, you will see that we can transfer the energy from photons to electrons, and then measure the energy of the electron. We can do tests to make sure that each electron is only absorbing the energy from a single photon, and we can do tests to make sure that all of the energy from the photon goes to the electron. And we can do tests to measure the wavelength of the photons. Therefore, we can indirectly get very accurate measurements of the energy that a photon has as a function of wavelength.

[mxplxxx (OP)]

That is correct. There is no good way to directly measure the energy of a single photon.

Why is that?

[mxplxxx (OP)]

But, if you go back and look at the answers that were given to you, you will see that we can transfer the energy from photons to electrons, and then measure the energy of the electron. We can do tests to make sure that each electron is only absorbing the energy from a single photon, and we can do tests to make sure that all of the energy from the photon goes to the electron. And we can do tests to measure the wavelength of the photons. Therefore, we can indirectly get very accurate measurements of the energy that a photon has as a function of wavelength.

This is an extremely small set of energies and circumstances to generalize  E=hf from.

[yor_on]

You should look at it the other way around, which is a very nice title for a book btw :)
First you will need to check up how Planck got to the formula, after that how Einstein used it to further develop the idea of a 'light quanta'. When you've done that you need to check if the experimental values gained from that formula fits the descriptions made by Planck. If those work then the formula describes it correctly.

That's the way I did it.

[yor_on]

I think this one is a pretty good start
https://www.quora.com/In-laymans-terms-how-did-Max-Planck-come-up-with-the-Plancks-constant-and-what-problem-does-it-solve

What I find fascinating about Planck is his imagination. He had a very long sight, with that formula just being part of it. He would have made a excellent SF writer, in the best sense. https://www.nature.com/news/2007/071220/full/news.2007.389.html

[alancalverd]

Since all photons in monochromatic light have the same energy (by definition of "monochromatic") then any measurement of the wavelength of that light (including x-ray photons) is a measure of the energy of a single photon.

There are some interesting special cases such as the Mossbauer effect that filters photons to a very narrow range of energies, or photonuclear reactions that have a definable threshold, such as pair production which can only occur at or above 1.02 MeV.

We use electron volts conventionally because the number is more convenient than joules. 1eV = 1.6 x 10^-19 J - same thing, just differently expressed.

Since the double slit diffraction experiment works with a single photon, and the diffraction pattern depends on the photon wavelength, it is a direct measure of λ for a single photon. f = c/λ is a simple matter of definition so it is a measure of f. We measure E as I suggested earlier, as the threshold voltage required to emit light from a LED or  characteristic radiation from an x-ray anode. So what is missing?

[evan_au]

Since the double slit diffraction experiment works with a single photon, and the diffraction pattern depends on the photon wavelength, it is a direct measure of λ for a single photon

I agree with Alan's answer to the original question.

I just have a quibble with the suggestion that the double slit experiment can directly measure the wavelength λ for a single photon.
- The probability that the photon will strike a particular place on the screen is certainly a function of wavelength
- But the probability is non-zero for a wide range of positions and a wide range of wavelengths
- So the fact that a single photon strikes a particular point on the screen cannot distinguish (say) λ from 1.5λ
- In particular, the maximum and minimum probability for λ is also a maximum and minimum for λ/2 (when the angle is small)

So I would say that the double-slit apparatus is a good way to determine the wavelength of many photons in monochromatic light. (end of quibble)
- The monochromatic nature of a light source can be checked by a number of methods, including:
- With a prism (producing a rainbow)
- With a diffraction grating
See: https://en.wikipedia.org/wiki/Diffraction_grating

The diffraction grating also allows direct calculation of the wavelength of a monochromatic light source, and is a way of producing monochromatic light, by selecting a single emission line from a gas-discharge lamp.
- You can also use highly monochromatic light sources, such as a distributed-feedback laser
See: https://en.wikipedia.org/wiki/Distributed_feedback_laser

[yor_on]

 Yes, a nice modern example that we use to prove a duality Alan, but it was introduced by Thomas Young in the beginning of 1800 to  argue for interference, aka a wave theory. I was thinking of it but to see how Planck got to his model I used the history. That first link is pretty good actually,  it has several nice links baked into it. It's interesting to read them and think about how we, ever so slowly, wandered  from Newtons 'corpuscles' to 'waves',  to then defining the modern definition which becomes a duality.  https://en.wikipedia.org/wiki/Young%27s_interference_experiment

[alancalverd]

Except that "duality" is not modern thinking!

Electromagnetic radiation can be modelled as particles or waves, but only human vanity (the antithesis of science) would pretend it is either or both. 

Anyway the statement that E_photon= hf is either an implicit statement of duality (a property of a particle  equals a property of a wave), or blindingly obvious since h is defined as E/f or Eλ/c (deBroglie's original concept).

Apologies for the italics, but these words are often confused in the minds of nonscientists - particularly politicians and economists.

[alancalverd]

I just have a quibble with the suggestion that the double slit experiment can directly measure the wavelength λ for a single photon.- The probability that the photon will strike a particular place on the screen is certainly a function of wavelength- But the probability is non-zero for a wide range of positions and a wide range of wavelengths- So the fact that a single photon strikes a particular point on the screen cannot distinguish (say) λ from 1.5λ- In particular, the maximum and minimum probability for λ is also a maximum and minimum for λ/2 (when the angle is small)

If you use a LED to replicate Taylor's double-slit single-photon experiment, there are no harmonics or subharmonics  of f or λ present. You can even add a quarter-wave filter to narrow the bandwidth.

Now here's what occurred to me in my schooldays. If a single photon can interfere with itself, then the exit photon must appear in several places simultaneously to produce the observed pattern. So if we input a 1-attojoule photon (somewhere in the violet end of the visible spectrum, I think - too early for mental arithmetic!) we can detect 1 attojoule in several places simultaneously on the exit side. We have generated energy  from  nothing! This either destroys the entire concept of classical duality, or makes the Big Bang inevitable: you can generate an infinite amount of energy (hence mass) from a single photon.

[mxplxxx (OP)]

Except that "duality" is not modern thinking!

According to who?

[alancalverd]

Anyone who deals with quantum mechanics, radiation measurement, or simple logic.

To take one practical example, x-ray crystallographers have been exploiting the diffraction of electromagnetic radiation to study the structures of materials and molecules for over 100 years, but for the last 50 years or so we have started counting photons electronically (rather than using photographic film) to determine the intensity of the diffracted beams. The idea that a photon knows in advance what we want it to do at a particular point in the apparatus is absurd, but appropriate application of the wave and particle models allows us to capture the data (from "particle-like" events) and process it (through continuous fourier transform wave equations) it to reveal the spatial disposition of atoms in the crystal.

At very low photon fluxes we can detect individual interaction ("single photon") events. Now if a "single photon" consists of a truncated "chirp" burst of waves,  Fourier tells us that we should be able to detect harmonics of the fundamental frequency. associated with, say, a LED or similar single spectral line. We can't.

[yor_on]

Don't really know when the idea of a 'duality' was first coined Alan, but to me it's recent anyway :)
=
Sorry, missed that mxplxxx and you already discussed it