# Difference between ionization energy and work function

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## Main Question or Discussion Point

I'm looking at the ionization energy of some elements and their work function. To my astonishment these energies are quite different (factor close to 2).
How is that possible that if say a piece of copper has almost all its atoms in the fundamental state (room temperature ensures this I guess), it costs less energy to remove an electron from any atom of the solid than if the atom was alone. It must be because the other atoms are somehow "helping you" to remove the electron. Hmm, and if electrons are "shared" since it's a good conductor wouldn't atoms be constantly ionized and getting back to neutral charge after a very quick time?
I'm somehow confused on what's going on.
Also I'm interested in knowing the work function of hydrogen molecule to see how it differs from the ionization energy of the H atom, namely 13.6 eV. I don't find this information on the internet.

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TeethWhitener
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How is that possible that if say a piece of copper has almost all its atoms in the fundamental state
“Almost” is doing some heavy lifting in this sentence. Yes, the work function of a metal differs from the ionization energy of a constituent atom because the electrons in the collection of atoms see a different potential than electrons in the individual atom. Another way to look at it is from a chemical point of view:
$$M \longrightarrow M^+ + e^-$$
Consider the ionized system. The positive charge left behind by the electron’s removal can spread out over a much larger area in a solid than in the atom. This stabilizes the solid versus the atom.
Hmm, and if electrons are "shared" since it's a good conductor wouldn't atoms be constantly ionized and getting back to neutral charge after a very quick time?
I don’t quite understand this. At finite temperature, there are always going to be some electrons above the Fermi level, but the charges are delocalized over the structure, so it’s not like individual atoms in the solid are being fully ionized.
Also I'm interested in knowing the work function of hydrogen molecule to see how it differs from the ionization energy of the H atom, namely 13.6 eV. I don't find this information on the internet.
The case is quite different for a small molecule like a diatomic. In the case of hydrogen, ##H_2## has a larger ionization energy than the atom. This is because the 1s atomic orbitals combine to form a sigma bonding molecular orbital that is lower in energy than the individual atoms, so the work required to remove an electron from the molecule is higher than the atom.

But this varies from species to species. The ionization energies of ##C_2## and ##N_2## are larger than the respective atomic species, but the ionization energies of ##O_2## and ##F_2## are lower than their respective atoms. This is because the ground states of ##O_2## and ##F_2## partially fill antibonding molecular orbitals, which are higher in energy than the atomic orbitals.

Why is the collective effect different for a solid than for a diatomic molecule? Essentially because the stabilizing charge delocalization that occurs in a solid still can’t occur much in a diatomic—it’s still too small. However, as the molecule grows in size, the delocalization effect can take over pretty quickly. Even for something as small as the allyl radical (##C_3H_5##) where delocalization is known to be important, the ionization energy is already less than that of a carbon atom.

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