We Have Lost a Half of Solar Energy, the Half that We Never Had
Experimental/Real Physicists Should Love It!
Addition to Classical Physics Beyond Einstein’s
4.3. Light Energy E = h×f/2 and Millikan’s Experiment
Let’s review Planck’s famous formula with his constant h for light/photon energy E = h×f, where f is the frequency of light/photon and h=6.626×10⁻³⁴ Joule×Sec constant. We have a strong reason for this revision because in the previous section we changed Einstein’s mass-energy formula E = m×c² to E = m×c²/2, and because by energy conservation principle, Einstein’s and Planck’s formulas are linked through electron-positron annihilation, when mass energy of particles is converted into light energy. There are various setups/configurations of electron-positron annihilation, with different numbers of photons at different energies output, but we are interested in the most popular case called para-positronium annihilation with two-photon outcome. The mass-energy of the electron (the same energy and mass the positron has) equals the energy of an annihilation photon. But now with Einstein’s formula revised to E = m×c²/2, we either have to change Planck’s formula to E = h×f/2 or change h-value to 3.313×10⁻³⁴ Joule×Sec. But before doing that, we have to review the experiments measuring light energy and h value.
The neatest setup, considered as an icon of experimental physics, is 1912–1915 Millikan’s test for photon energy: he measured energy of light by measuring energy of electrons kicked out by photoelectric effect from a metallic plate, and he did that by measuring stop-voltage preventing electrons from reaching another metallic plate:
For more detailed explanations of Millikan’s experiment, please watch the video above and/or the video below.
Electric Current versus Voltage graph below shows that near stop-voltage current is really small: there are very few electrons having such high energy (energy of these few electrons is proportional to the stop-voltage):
By measuring stop-voltage, Millikan measured maximal kinetic energy (which is m×v²/2, where m is mass of the electron and v its velocity) of photoelectrons, thus, he measured maximum velocity of these photoelectrons. And the graph clearly shows that the population of such top-velocity electrons is extremely small. Millikan assumed that energy of the incident photons equals the energy of the fastest electrons. Is it true or is there twice the difference between these energies? I pay attention to this part specifically, because by-the-factor-of-2 mistakes are quite common in energy and velocity estimates, take for example Feynman’s “(Wrong!)” comment in his “43–3 The drift speed” lecture at https://www.feynmanlectures.caltech.edu/I_43.html. Now, let’s consider collision at 90° angle between two electrons, initially having the same velocity v:
The pink ball stops after collision, and the blue ball gains velocity, and its energy becomes twice higher than before the collision: m × (√2 × v)² / 2 = m × v² = 2×(m×v²/2). Both total energy and total momentum are conserved in such elastic collisions. Applying that to photoelectrons: besides the main population of electrons having about the same velocity v (pointing in various directions), there is a very small population of collided electrons, where each electron has up to doubled energy than an electron from the main population of electrons has. Here is Millikan’s mistake: he put an equal sign between photons energy and the maximum of electron’s energy, when he should have used only half of the max of electron energy. We just have corrected Planck’s photon energy formula to E = h×f/2, or if you still prefer E = h×f, then you have to change the value of h from 6.626×10⁻³⁴ to 3.313×10⁻³⁴ Joule×Sec. But changing value of h is more impactful for physics, as you will have to check all formulas in physics that include h.
P.S. I just added this section to my 📖 book, which is available in PDF, on Amazon and Google Books.
