# Experimental verification for the de-Broglie - Einstein relation $E= h \nu$ for a particle

Is there any experiment that confirms the relation $E=h \nu$ for material(I mean the matter waves) objects? I mean, from the photo electric effect experiment, we came to know that photon carries energy $E= h\nu$ for a given frequency, but how did they verify it for a particle ?

• en.wikipedia.org/wiki/Davisson%E2%80%93Germer_experiment – hyportnex Jul 2 '17 at 16:44
• @hyportnex There's an argument to be made that Davisson-Germer interference experiments only measure the wavelength, not the frequency, of matter waves. Akaash points specifically to the de Broglie-Einstein relation, which to my mind implies that they're asking for the $E=h\nu$ connection, not the pure de Broglie $p=h/\lambda$ version, which is the one that involves the wavelength. – Emilio Pisanty Jul 2 '17 at 19:16

The most striking experiment [1] I know of consist in diffracting fullerenes molecules, formula $\text{C}_{60}$, the famous buckyballs. The de Broglie wavelength is about 400 times smaller than the actual diameter of the molecule. Here is how the calculation goes: the average speed of the molecules was $v=220\ \mathrm{m/s}$, determined by a time-of-flight method; the mass of one molecule is $m=1.2\times 10^{-21}\ \mathrm{g}$, resulting in a momentum $p=\gamma m v\approx mv=2.6\times 10^{-19}\ \mathrm{gm/s}$, and therefore in a de Broglie wavelength $\lambda=h/p=2.5\times 10^{-12}\ \mathrm{m}$. From that value, one can predict the diffraction pattern and check it compares well with the experimental measurement, which it does according to the article.

What about the other de Broglie relation you were interested in then? $E=\gamma m c^2$ but $\gamma\approx 1$ here (I already used that in the computation of the momentum $p$), i.e. only the rest energy matters, and we get $E=6.7\times 10^{11}\ \mathrm{eV}$, which is way higher than the electromagnetic binding energies, so we are ok with just using the bare mass of the 60 Carbon atoms. That gives $\nu=1.6\times 10^{26}\ \mathrm{Hz}$. But none of that was measured as the diffraction patterns depends on the wavelength.

Moreover this molecule is a very complex object with many internal degrees of vibrations and a lot of very close energy levels. Thus it can be argued that this is as close to a classical object as we can get for this type of experiment.

[1] Markus Arndt, Olaf Nairz, Julian Vos-Andreae, Claudia Keller, Gerbrand van der Zouw, and Anton Zeilinger, Wave-particle duality of c60 molecules, Nature 401 (1999), 680--682

• So from the above procedure I get the momentum and hence the wavelength from the de-Broglie relation. So we can calculate the energy(by using E=hv). So the obtained value of E must tally with the experimental value of the energy right? How was this done? E=hv gives me an energy of 7.956 10^{-14} . But if I use 0.5 mv^2 , I get a different answer. – Akaash Srikanth Jul 2 '17 at 18:57
• I got it till the part where you obtain the frequency. But after this if i again use the relation c/v(frequency) , I get a different wavelength than the one mentioned above (2.5 * 10^-12 ). – Akaash Srikanth Jul 2 '17 at 19:26
• That relation is only valid for a wave propagating at the speed of light, which is not the case here. The equivalent relation would be $\lambda\nu=E/p$. – user154997 Jul 2 '17 at 19:32
• Yes,I got it now. So basically, we find the calculated wavelength agrees with the value of the wavelength obtained experimentally. And then we use "Debroglie-Einstein" relation to calculate the energy from which we calculate the frequency. So now we check whether the frequency and wavelength agree with each other through the above mentioned equation. Did I get that right? – Akaash Srikanth Jul 3 '17 at 10:19
• Well, yes, but note that $p=h/\lambda$ and $E=h\nu$ imply $\lambda\nu=E/p$. – user154997 Jul 3 '17 at 10:23

The clearest examples of this are various forms of the photoelectric effect, either in ionization from solids, ionization of atoms and molecules in the gas phase, or in excitation and de-excitation of bound-bound transitions in the gas phase.

If you buy into the matter-wave hypothesis from the beginning, then it is clear that these phenomena couple the frequency of the matter wave to that of the incoming electromagnetic radiation. There's a number of ways you can do this, depending on exactly what you're looking for, but they all play with the same set of fundamental ideas.

As an example, you can take a sample of hydrogen atoms at rest and bombard them with electrons with a narrowband tunable kinetic energy, and then observe for fluorescence from the gas; you will find that there are certain specific regions of the bombardment energy that produce a much higher fluorescence, and that each of them produces a different colour in the fluorescence.

It is always a bit tricky to distinguish between experiments that couple to the wavelength (like the double-slit interference experiments in Luc and anna's answers) and experiments that couple directly to the frequency, but for experiments in the gas phase it is very hard to see how the wavelength of the light can be relevant for the coupling (since it's much longer than the size of each atom) so the only way the atomic dynamics can couple to the light is through its frequency.

Here, if you buy into wave mechanics, the connection is clear: you have formed a superposition of atomic states which oscillate at different frequencies, and their interference couples to the EM field, to a first approximation as I described here.

If you don't really buy the wave mechanics and you're looking for evidence to convince you, then it gets a bit tricky - ultimately, the reason we buy the wave mechanics isn't because of a single experiment but because of the full edifices of experiment and theory and how they interact with each other - but you nevertheless have a phenomenon that (i) clearly depends on the supplied mechanical energy, (ii) is clearly resonant, as a confined wave would, together with a Lorentzian lineshape if your experiment is precise enough, and (iii) its effects wind up imprinted in the temporal frequency of an external system.

That the electron trajectories obey a wave probability is shown clearly in this accumulation of single electrons scattering on two slits.

Results of a double-slit-experiment performed by Dr. Tonomura showing the build-up of an interference pattern of single electrons. Numbers of electrons are 11 (a), 200 (b), 6000 (c), 40000 (d), 140000 (e).

From top to bottom, where the footprint of the single electrons are detected , the accumulation of the distribution of the scattered electrons ( a probability distribution) shows the interference pattern of wave nature.

The wavelength is consistent with the quantum mechanical predictions. (original paper copies in pdf exist when searching title and author , where the experiment is described and the conclusions drawn.

• Can't an argument be made that the interference pattern is set by synchrotron radiation from the electrons? The radiation between the slit and detection screen form the familiar interference pattern on the screen but is not seen as visible light. Then as the electrons travels through the slit their trajectories are continuously affected by the reflecting radiation set in the pattern. The electrons are guided and corralled into certain location and on impact create more synchrotron radiation which reflects back-and-forth between the walls maintaining the original photon fringe pattern. – Bill Alsept Jul 2 '17 at 17:50
• @BillAlsept There is no synchrotron radiation, only a single scatter from the slits. The word "synchrotron itsself denotes multiple scatters or radial acceleration: en.wikipedia.org/wiki/Synchrotron_radiationReflecting from what? – anna v Jul 2 '17 at 17:54
• I thought accelerated electrons would emit radiation – Bill Alsept Jul 2 '17 at 17:56
• These electrons are a beam with no acceleration. And anyway they come one by one, a very weak beam. – anna v Jul 2 '17 at 17:57
• The above image is the result of exposing the screen to electron beams for a very long time right? So we get the wavelength and try matching it with our calculated debroglie wavelength to verify the result. But how do we match the experimental energy of the electron and the theoretical value of the electron(E=hv) ? – Akaash Srikanth Jul 2 '17 at 19:06