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In atoms, only certain orbitals are permitted (quantization)- To calculate these orbitals, you need to solve the wave equation for the electron.- Some of these orbitals are spherical, but others look like a cluster of balloons assembled by a clown. How do you calculate the radius and centripetal motion for these?If you want a simple understanding, have a look at Bohr's model of the atom, where an electron's angular momentum is quantized (classical physics has no equivalent).- Or de Broglie's model where the electron has a wavelength, and that wavelength must have an integer number of wavelengths to be stable (classical physics has no equivalent).- But for a good model, you have to solve the relativistic Schroedinger equation, which gets quite complex for anything bigger than a hydrogen atom. Even a Hydrogen atom is beyond what they are paying me here!See: https://en.wikipedia.org/wiki/Atomic_orbital#Bohr_atom
The Visible Spectrum of Hydrogen vs. DeuteriumSuzanne Fiacco '01Abstract: The purpose of this project is to use a reflection spectrometer to find the differences in wavelengths between the spectrum of hydrogen atoms and the spectrum of deuterium atoms. Hydrogen and deuterium share common characteristics. Deuterium is also known as heavy hydrogen because the weight of deuterium is twice that of hydrogen. Hydrogen is the simplest atom, which consists of one proton and one electron while deuterium is made up of one neutron, one proton, and one electron. Since the physical properties indicate that hydrogen and deuterium are very similar, one would expect their wavelengths to be very similar. In this projects, we calculated three of the visible wavelengths in the hydrogen spectrum to be 656.478 nm, 486.542 nm, and 434.415 nm. For deuterium we calculated that these wavelengths shift to 656.296 nm, 486.409 nm, and 434.295 nm respectively due to the additional mass in the neutron in the nucleus. In doing this experiment we measured the wavelengths in the hydrogen atom to be 656.489 nm, 486.44 nm, and 434.238 nm. For deuterium we measured the visible wavelengths to be 656.295 nm, 486.315 nm, and 434.115 nm. After measuring the intensity verses wavelength for the visible spectrum, we can determine the shift in wavelength for the red, blue, and violet lines as we change the source from hydrogen to deuterium. We conclude that the percent error between the differences in wavelengths between the spectrum of hydrogen and the spectrum of deuterium to be 1.65%, 5.3%, and 2.5% for the wavelengths of red, blue, and violet respectively.For more information, contact Dr. Catherine Jahncke: cjah@stlawu.edu
Maxwell's equations work perfectly well within their limited scope- they are not meant to be a TOE. I think I have already pointed this out.
What does Maxwellian electromagnetism actually predict in each of those experiments?
Quote from: hamdani yusuf on 31/05/2023 09:47:27What does Maxwellian electromagnetism actually predict in each of those experiments? Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less. The lift equation F = 0.5ρAv2 tells you how an aircraft will fly, but nothing about tyre skid on landing. So what?
Maxwell's equations describe the propagation of electromagnetic radiation, nothing more or less.
The lift equation F = 0.5ρAv2 tells you how an aircraft will fly, but nothing about tyre skid on landing. So what?
Quote from: paul cotter on 31/05/2023 10:43:26Maxwell's equations work perfectly well within their limited scope- they are not meant to be a TOE. I think I have already pointed this out.Don't you want to know what makes it not working in microscopic scale? Or rather, what makes it work in most macroscopic scale? Note that flat earth model and Geocentric model also work perfectly well within their limited scope- they are not meant to be a TOE.
They don't, for example, explain the blue sky.
Quote from: Bored chemist on 31/05/2023 12:52:47They don't, for example, explain the blue sky.Rayleigh scattering in the atmosphere is a classical continuum effect, entirely consistent with Maxwell.
Nobody said they do.
I seem to have to keep repeating the bloody obvious: Maxwell's equations describe the propagation of EM radiation,
Maxwell's equations only describe the propagation of EM radiation in a vacuum.
No. They apply to any medium if you substitute εm and μm for ε0 and μ0.
To make life easy, we measure and publish dimensionless relative permittivities and permeabilities for various materials (including air and metamaterials) so you can just multiply the vacuum value as appropriate.
If you get a high enough field strength practically everything behaves non-linearly.Even air misbehaves if you try hard enough.//www.youtube.com/watch?v=Z1Xky_ermd4
Quote from: Bored chemist on 02/06/2023 18:09:20If you get a high enough field strength practically everything behaves non-linearly.Even air misbehaves if you try hard enough.//www.youtube.com/watch?v=Z1Xky_ermd4I'm wondering how it would look like under different air pressure.