From my understanding, the fermi level of a metal at room temperature indicates an energy level at which 50% of electrons exist above, and 50% exist below. Why then, is the fermi level in UPS spectra of metals observed as a distinct cutoff point? It seems like it should be continuous.
Hello, I have no idea on how to do this problem. I'd really appreciate someone showing me how to do this, you can change the element/numbers. Thanks in advance.
Here's what's being asked -
Using photoelectron spectroscopy, the ionization energy of the least tightly bound valence electron on Cl was determined to be 13.0 eV.
Calculate the Zeffective for this electron
Hello. I was wondering why on an XPS graph (with # emitted electrons vs the binding energy) why there seems to be emitted electrons that don't correspond exactly to an electron energy shell level.
This wikipedia article has some neat images I refer to:
https://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy
In the second image labelled "XPS physics - the photoelectric effect*.",* you see a graph with peaks corresponding to particular electron shells, but there's also "mounds" that these peaks rest on.
I was really curious why the mounds seemed to go up and down. Do these mounds correspond to auger electron emissions (i.e "leftover" energy is going into these electrons after a higher energy electron is released and the core shell is filled again?)
Thanks for any response to my perhaps naive question!
Journal of the American Chemical SocietyDOI: 10.1021/jacs.0c06508
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https://preview.redd.it/lbsbbgampem41.jpg?width=700&format=pjpg&auto=webp&s=cd06cb5255966c5d90fdfc93b1148c932bbdb223
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... keep reading on reddit ➡There is an old video that was on YouTube of an MIT professor explaining PES and it's since been removed. It was a great explanation and now it's gone! All I remember is that he had about 9 chalkboards going at once, moving them up and down, and he draw a picture of the PES device. I'd love to have this file.
This is my understanding of it: A light (of energy hv) is shined onto a sample of gaseous atoms, which will absorb a certain amount of light (the ionization energy). The difference between the inputted light energy hv and the ionization energy is the kinetic energy that will be released with the electrons.
What I'm not as sure of: Scientists control the voltage of the analyzer (charged magnetic plates?) until the electrons hit the detector and they can measure the kinetic energy of the electrons. This would mean that in the formula hv=IE+KE, they know hv initially (since they control the amount of light?), they test for KE, and they calculate IE... Is this correct?
Edit: saw "photoelectron count rate" on the y-axis of a graph and KE (ev) on the x-axis and now I'm more confused- this means that they control the KE, but how does the # e-/sec give the IE of the gaseous atom?
There may be several answers, but generally? And what elements are used to create this source? Thanks.
Being fairly optimistic, I hope that someone here could actually help me.
Recently I came across some publications' experimental data that can actually fit my calculations. The catch is, that (I think) the photoelectron spectrum of the gold surface was not corrected for at all. I have no clue whatsoever on how to even start correcting for it, and if I do not correct for it I cannot really say if my calculations are on the spot.
Any one with experience care to point me in any direction?
For background, I do photography as a hobby, teach myself technical material, look for problems to solve, and brainstorm ideas for solutions.
Problem
Most people stick with their smartphones on the basis that the best camera is what you actually use. The problem is that while they've improved, they're still terrible in low light. People want to photograph both subjects with broad ratios between bright and dark as well as scenes that are more moderate or low in contrast ratio.
With video on a bright day, even moderate sensitivity levels like ISO 50 can be awkward because of wide aperture lenses. Under the "Sunny 16" rule, a wide aperture is difficult without low sensitivity and high well capacity to compensate.
With candlelight scenes with flames and areas reflecting off flames, high noise and blown out highlights are a given. With moonless nights lit only stars and no light pollution, even the ISO 4 million camera from Canon will scrape by. Inability to capture starry moonless on smartphones is a given.
If fewer photons are collected but the read noise is the same, low light performance and dynamic range can degrade.
Ideas
There are solutions I looked at for this challenge. They are:
- OPF CMOS. If I understand correctly, the voltage gets adjusted as a way of adjusting sensitivity. The one Panasonic is working on has a "high" full well capacity. How does this compare to avalanche gain?
- QIS CMOS. With combining single bit values for picture construction, how can it handle bursts from strobes and xenon flash units? How can the processor handle the high volume of data generated compared to regular CMOS (if it does create a high volume of data)?
What're your thoughts?
