If metals experience the photoelectric effect, then why aren't they used directly in solar power generation, as opposed to expensive silicon cells?
A better question would be, why is the quantum yield of metals in general quite low?
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.Sign up to join this community
There might be a concise and elegant answer by principle but this is what comes to my mind in a somehow sparse but interconnected order.
First, the photoelectric effect does not establish a difference of potential nor the metal would sustain it in its bulk. This means that photoemission occurs in random directions and electrons are not flowing as current. Thus, the external circuit must contain not only the load but also a source of potential difference. This is the arrangement in applications such as phototubes, photomultipliers, ...
Photoelectric effect involves the surface only. At least at energy relevant to solar energy conversion on Earth. While this is irrilevant for measuring a small number of photons, it makes the "practical absorption" of a metal solar cell very small. Also consider that metals are reflecting the visible light. You can group this as being the "quantum efficiency" of the cell as thought per number of impinging photons. It has to be low.
Metals with the lower work functions are inherently unstable in atmosphere. Beside degradation of the bulk electrode, thin layer of oxides already would prevent electron emission. Atmosphere hinders emission in itself, so anyway everything must be packaged under vacuum. Again this is viable for a bulb but not for an extended surface to be placed on a roof.