Question

Can you explains why the alkali metals differ in their photoelectric sensitiveness in a perfectly definite order, which is that of their degree of electro-positiveness, as shown by their position in the periodic table of the elements??

Answer 1

any idea @Carl_Pham ??

Answer 2

Electropositiveness is not the right criterion, ionization potential is, because a photoelectric response is generated when a photon ejects an electron from a metal, which is related to how easy it is to pry a valence electron out of the atom. As you go down or to the left in the Periodic Table, the effective electrostatic force holding the outermost electrons in place decreases, and the ionization potential drops, making for easier ejection of photoelectrons.

It is also true that electropositivity -- how readily the element gives up control of its valence electrons in a bonding situation -- also increases, and for the same basic reason: these atoms lose control of their valence electrons more easily.

Answer 3

ok.., anyting else???

Answer 4

For example, consider the IP of the Group 2A metals (kJ/mol):

Be: 900
Mg 740
Ca: 590
Sr: 550
Ba: 500

And the work functions (eV):

Be: 5.0
Mg: 3.7
Ca: 2.9
Sr: 2.6
Ba: 2.5

Answer 5

Sure, there's other stuff. The nature of the metal lattice is important, too, because that determines details of the band structure. Keep in mind you're NOT actually talking about ripping an electron off individual atoms, but rather ejecting one from the band structure of the solid. The band structure is related to the energy level structure of the atom, but it's not exactly the same, because the interaction between the atoms is governed by their distance from each other and arrangement in the solid.

So there's a correlation between the electronic properties of the atoms, in this case IP, and the electronic properties of the solid metal, in this case work function, but it's not always 1-to-1. There are details of the solid structure that matter. How exactly they do is fairly complex matter. I'm not that familiar with it. You'd have to dig into a solid-state physics book, e.g. Ashcroft and Mermin.

Answer 6

thank you so much for you explain.., can you help me again.????

Answer 7

@Carl_Pham