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by lukko
61 days ago
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Ah amazing - thank you for the response! I have a couple of related questions - is it that the non 2 pi frequencies exist, but they destructively interfere so we can't see them? My understanding is that the radial function for the electron is zero at the nucleus - there is no possibility of it being found there - but why is that the case? |
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In an atom, angular wavefunctions with wavelengths non-integer divisions of 2pi can’t exist because of the boundary conditions on the wave equation. A free electron can have any wavelength, but once you put it in a box (confine it to the potential around a proton in a Hydrogen atom) the non-integer wavelengths aren’t allowed
I think it’s instructive to think about what the wavefunction represents. It’s square is the electron probability density (technically the wavefunction is complex valued so it’s the wavefunction times it’s complex conjugate). If you have a non-integer multiple wavelength then the wavefunction goes out of phase with its complex conjugate after one period, and if you integrate over the angular domain the electron probability has to be zero everywhere.
This also answers your second question. The radial solution to the wave equation for hydrogen gives you the Laguerre polynomials. They don’t all go to zero at the nucleus though, actually the first one has a maximum at zero because it scales like exp(-r) (See fig 4.10.2 on chem.libretexts linked below). But when you do a volume integral to calculate the electron probability, the probability near the nucleus is low because the integration volume is small even though the wavefunction is large
https://en.wikipedia.org/wiki/Laguerre_polynomials
https://chem.libretexts.org/Courses/University_of_California...