Work function and ionization energy

AI Thread Summary
The work function, which is the energy needed to remove an electron from a metal surface, is generally about half the ionization energy required to remove an electron from a free atom of the same metal. This difference is attributed to the presence of free electrons in the metal lattice, making it easier to extract an electron compared to the more tightly bound electrons in a free atom. The metallic bonds allow for electron sharing, leading to collective effects such as hybridization and the formation of conduction and valence bands. There is no specific formula directly relating work function and ionization energy, and the "approximately half" value lacks significant meaning. Overall, the properties of solids differ fundamentally from those of isolated atoms due to these collective interactions.
tenchotomic
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Why is it that the energy required to pull an electron out of a metal surface(work function) is approximately half of the energy required to pull an electron out of the free atom(ionization energy) of the same metal (or element)?
Is there any formula relating the two quantities?
 
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I think it is because in a metal lattice there are already free electrons moving around everywhere. Pulling one of these out is much easier than a free atom which has it's electrons bound. In effect, the metallic bonds make it easier because they "share" electrons.
 
tenchotomic said:
Why is it that the energy required to pull an electron out of a metal surface(work function) is approximately half of the energy required to pull an electron out of the free atom(ionization energy) of the same metal (or element)?
Is there any formula relating the two quantities?

This is exactly the reason why a solid is not the same as an isolated atom. The formation of a solid means that the valence shell of the atoms have overlapped with more than one other atom, causing hybridization, etc. The formation of conduction band, valence band, band gap, etc. are all COLLECTIVE effects due to the all the atoms, not just one.

BTW, there's no significance to the "approximately half" value of the work function.

Zz.
 

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