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    If the incidence light acts like a wave rather than a particle, the energy of the wave would be absorbed by the metal. I am confusing in here... Would all energy be absorbed by the metal, or only some of it then the remaining energy hit the electron?
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    (Original post by alesha98)
    If the incidence light acts like a wave rather than a particle, the energy of the wave would be absorbed by the metal. I am confusing in here... Would all energy be absorbed by the metal, or only some of it then the remaining energy hit the electron?
    Surely the photoelectric has light behaving as particles rather than a wave?
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    (Original post by Zacken)
    Surely the photoelectric has light behaving as particles rather than a wave?
    Yes you are right, but in my question, i am asking If the incidence light acts like a wave, what would happen to the energy?
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    (Original post by alesha98)
    If the incidence light acts like a wave rather than a particle, the energy of the wave would be absorbed by the metal. I am confusing in here... Would all energy be absorbed by the metal, or only some of it then the remaining energy hit the electron?
    Hmmm, you don't really need to go that far in depth, you just need to know that if light was a wave then eventually electrons could be released from the surface of the metal but that's not what happens.
    Shine light on a metal and it'll never have a current flowing through it. The light must be in the UV range. Besides this i'll try and answer your question.

    If it's a wave then i'm assuming that all of the wave is sorta spread out. So some of it hits the metal atoms and some hits the electrons. In anycase the light isn't high enough frequency to reach the work function

    (Original post by Zacken)
    Surely the photoelectric has light behaving as particles rather than a wave?
    It acts as both

    eyyyyy, it's just one of those physics things...
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    (Original post by alesha98)
    Yes you are right, but in my question, i am asking If the incidence light acts like a wave, what would happen to the energy?
    The energy of each photon (light acting as a particle) is absorbed by electrons which allows them to be liberated out of the metal- the energy is then conserved into kinetic energy in the electrons as they create a current through the circuit.
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    (Original post by thefatone)

    It acts as both

    eyyyyy, it's just one of those physics things...
    I know that is acts as both in general, but for the photoelectric effect, it only acts as particles, doesn't it?
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    (Original post by Zacken)
    I know that is acts as both in general, but for the photoelectric effect, it only acts as particles, doesn't it?
    Yes photoelectic effect is proof that light acts as a particle
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    (Original post by thefatone)
    Yes photoelectic effect is proof that light acts as a particle
    Yeah, that's what I was saying.
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    (Original post by alesha98)
    If the incidence light acts like a wave rather than a particle, the energy of the wave would be absorbed by the metal. I am confusing in here... Would all energy be absorbed by the metal, or only some of it then the remaining energy hit the electron?
    For EM waves incident upon a metal the waves only penetrate to the "skin depth" of the metal as the wavevector becomes complex. This is why metals are shiny, the transmitted wave decays exponentially (as e^{-z/\delta}}) and almost all the energy that's put in comes back out
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    (Original post by langlitz)
    For EM waves incident upon a metal the waves only penetrate to the "skin depth" of the metal as the wavevector becomes complex. This is why metals are shiny, the transmitted wave decays exponentially (as e^{-z/\delta}}) and almost all the energy that's put in comes back out
    Thankyou, this is what I want

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    (Original post by alesha98)
    Thankyou, this is what I want

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    You're welcome
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    (Original post by alesha98)
    If the incidence light acts like a wave rather than a particle, the energy of the wave would be absorbed by the metal. I am confusing in here... Would all energy be absorbed by the metal, or only some of it then the remaining energy hit the electron?
    The photoelectric effect is used to demonstrate how light acts as a particle, hence why an electron is emitted for each photon which is incident at the metal surface (given that eh photon energ >= work function).
 
 
 
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