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Why is the entire 3p sublevel peak shifted to the right on a PES of Sulfur vs Phosphorus?
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Kim Aaron
Why is the entire 3p sublevel peak shifted to the right on a PES of Sulfur vs Phosphorus?
My question was rooted in a misunderstanding with the process of Photoelectron spectroscopy.
Originally, I thought that photons were beamed until all electrons were lost from an atom, which would report the first, second, third, etc. ionization energy, where the second ionization energy starts with the +1 cation, and the third with the +2 cation, etc.
However, Photoelectron spectroscopy actually finds the binding energy for each electron in the ground state. The reason it can do this is that:
it uses a photon energy that can knock off any electron in the atoms, and
since there is a large amounts of atoms and finite photons, very few atoms get struck by a photon and lose even a single electron, so it would be even more rare for an atom that already lost an electron to be struck by a photon and lose another electron.
For others struggling with this concept, the answer here does a good job of explaining the difference between first ionization energy and binding energy.
Here is a Khan Academy link that explains the process of Photoelectron spectroscopy in simple terms.
Thank you to @Karsten for helping with this question!
My question was rooted in a misunderstanding with the process of Photoelectron spectroscopy.
Originally, I thought that photons were beamed until all electrons were lost from an atom, which would report the first, second, third, etc. ionization energy, where the second ionization energy starts with the +1 cation, and the third with the +2 cation, etc.
However, Photoelectron spectroscopy actually finds the binding energy for each electron in the ground state. The reason it can do this is that:
it uses a photon energy that can knock off any electron in the atoms, and
since there is a large amounts of atoms and finite photons, very few atoms get struck by a photon and lose even a single electron, so it would be even more rare for an atom that already lost an electron to be struck by a photon and lose another electron.
For others struggling with this concept, the answer here does a good job of explaining the difference between first ionization energy and binding energy.Here is a Khan Academy link that explains the process of Photoelectron spectroscopy in simple terms.
Thank you to @Karsten for helping with this question!
My question was rooted in a misunderstanding with the process of Photoelectron spectroscopy.
Originally, I thought that photons were beamed until all electrons were lost from an atom, which would report the first, second, third, etc. ionization energy, where the second ionization energy starts with the +1 cation, and the third with the +2 cation, etc.
However, Photoelectron spectroscopy actually finds the binding energy for each electron in the ground state. The reason it can do this is that:
For others struggling with this concept, the answer here does a good job of explaining the difference between first ionization energy and binding energy. Here is a Khan Academy link that explains the process of Photoelectron spectroscopy in simple terms.
Thank you to @Karsten for helping with this question!
My question was rooted in a misunderstanding with the process of Photoelectron spectroscopy.
Originally, I thought that photons were beamed until all electrons were lost from an atom, which would report the first, second, third, etc. ionization energy, where the second ionization energy starts with the +1 cation, and the third with the +2 cation, etc.
However, Photoelectron spectroscopy actually finds the binding energy for each electron in the ground state. The reason it can do this is that:
For others struggling with this concept, the answer here does a good job of explaining the difference between first ionization energy and binding energy.Here is a Khan Academy link that explains the process of Photoelectron spectroscopy in simple terms.
Thank you to @Karsten for helping with this question!
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