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I know from the book that electrons will be kicked out from the metal plate if the light of appropriate wavelength is exposed to the metal plate.

My mental model says if we let the light expose the plate in a long enough interval, the plate will have no electrons anymore. So what happens on the plate? Does it still looks like a plate as the original? Or it become lighter due to loss of electrons?

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Well, it will become lighter with each electron removed, but unless you do it in the vacuum, it will get its electrons back from the environment.

Also, unless we are talking about really high energies, only the weakly bound valence electrons will be removed. The work to remove the electron, $W_a$, becomes higher the more electrons have left.

Okay, now if we suppose that you have somehow managed to remove all of the valence electrons, then sure, the plate will behave differently, as the optical and electrical properties are related to the electronic structure. If you have kicked out all (quasi-)free electrons, it might not be such a good conductor any more, for example.

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Interesting explanation. Thank you @Lagerbaer. I don't know whether or not I have a chance to do an experiment like this. It looks so imaginative. :-) –  xport Dec 7 '10 at 5:03
You'll need a good vacuum, and a kathode/anode (I always mix these up...) that sucks away all the photo-electrons. And you need light of a high frequency. –  Lagerbaer Dec 7 '10 at 5:04
Electrons are so light (in the mass sense) that I can't imagine you'd be able to remove enough of them to make a measurable difference in the weight. At least not before the charge on the plate pulls more electrons in to replace the missing ones. –  David Z Dec 7 '10 at 5:35
Doesn't the energy required to kick off an electron increase as the charge increases? –  endolith Dec 8 '10 at 5:43
Yes. That's what I wrote above: "The work to remove the electron becomes higher the more electrons have left". –  Lagerbaer Dec 8 '10 at 6:49
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