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Introductory Physics Homework Help
Calculating Electron Emission and Reflection in the Photoelectric Effect
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[QUOTE="Zeynaz, post: 6164583, member: 660503"] [B]Homework Statement:[/B] The intensity of the light that falls onto the cathode is 6.0 W/m2. The area of the cathode of the photocell is 3.5 cm^2 b) Calculate the percentage of the photons incident on the cathode that release an electron. c) what happens to the energy of the photons that do not release an electron [B]Relevant Equations:[/B] E=hf V=IR P=IV I=n(q-electron) [ATTACH type="full" alt="241969"]241969[/ATTACH] The full questions is in the picture. I already solved a) and found 5.6E14 electrons per second For b) i first found the power of the light but just multiplying the intensity with the area: (6.0 W/m2)(3.5E-4 m^2) = 0.0021 W Then I tried to use the voltage from the graph but i am not sure which value i should use because i don't know the voltage that corresponds to the intensity given. So could you show me a way that can help me work it out? After finding the number of electrons that the light releases i would calculate the percetage by dividing the value in question a) by the value i find and multiply it with 100. (The correct answer should be 13% but i don't know how to get there) For c) i said that because the energy of the photons are not enough to eject the electrons on the plate are a reflected back. (But I am not sure if that's a correct statement) [/QUOTE]
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Introductory Physics Homework Help
Calculating Electron Emission and Reflection in the Photoelectric Effect
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