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The electric field of the "battery" allows you to measure the kinetic energy of the electrons, which is crucial for the understanding of the photoelectric effect, since that energy varies with the frequency of the light.


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As ACuriousMind says "The photoelectric effect is the observation that many metals emit electrons when light shines upon them." You don't need anything except the light and one plate for the photoelectric effect. Why are the other parts there? You presumably want to measure some things to understand the effect. Having the circuit allows you to measure ...


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"If a photon hits an electron". If only it were that simple! The electrons (in this case) are bound in atoms (or ions). If the photon has the "right" energy, then the probability of the atom making a transition from one state to another may become appreciable and the transition may occur. If the photon does not have the "right" energy - and you can think of ...


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The equation you cite is the energy of an electron that is produced as a result of the photoelectric effect. Your metal has a work function ($\phi$), and at minimum you require that energy, as supplied by a photon, to produce an electron. To produce any electrons at all, your source of photons would have to have an energy greater than the work function. ...


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The photoelectric effect involves the shell (electrons) of the atom. Photon energies are typically in the range of electron volts. To excite the nucleus, you'll need higher energies (MeV or hundreds of keV). What you are probably looking for is called photonuclear reaction. Anyways, so if an atom has too much energy and ejects an electron, is that ...


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It's a matter of definition. For example, by photon, do you mean first or second quantization (the latter being the canonical treatment)? Although there's no ambiguity in theoretical physics, the nature of photon has generated much debate in the community of applied physics. See, for example, these articles: http://arxiv.org/pdf/quant-ph/0605102.pdf ...


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You have really figured out the answer yourself. On one hand, you have a natural phenomenon (light) and on the other you have our models (wave description, photon description). When speaking of the reflection of rarefaction of photons, the author implicitly assumes the reader to know about the wave-particle duality (see links in the comments). I think you ...



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