Understanding the Photoelectric Effect and its Effects on Metal Decay

AI Thread Summary
The discussion centers on the photoelectric effect and its implications for metal decay. A positively charged metal plate illuminated with visible light does not lose its charge because the energy of visible light is insufficient to free electrons from the metal. Conversely, a negatively charged plate exposed to UV light does not become positively charged, as the photoelectric effect releases electrons rather than replenishing them. Concerns about whether the photoelectric effect causes metal decay arise, with the understanding that emitted electrons are not replaced, potentially leading to degradation over time. Overall, while the photoelectric effect facilitates electron emission, it does not directly contribute to the decay of metals in the short term.
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Im trying to get my head around a question in a book as i thought it would be good to give them ago as I am struggling.

The question is:

Explain why a metal plate does not lose its charge when;

a) the plate is + charged and illuminated with visible light.

b) the plate is - charged and illuminated with UV light

Also, when given wavelength and power of a wave, how do you work out the number of photons emitted per second?
 
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a)Even though the plate are being exposed for visible light, the light has not enough energy to free electrons from the metal.

b)The plate has already a negative charge, and won't become positive if you expose it to UV light. The effect releases electrons, its not filling up whit them.
 
I have a question about the photoelectric effect. Does it cause metal to decay? It involves emission of electrons from the metal and as far as I know there are no electrons being replaced: does the light cause the metal to break down after a long period of time?
 
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