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The relation is hf=1/2 mv^2 + W You know v(the velocity)and m(mass of an electron) and also you know the workfunction w The work function is the minimum energy that must be given to an electron to liberate it from the surface of a particular substance. In the photoelectric effect, electron excitationis achieved by absorption of a photon. If the photon's ...


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By Einstein's photoelectric effect $hv - hv_0 = 1/2mv^2 $ where $hv_0 $= work function and $hv$ is the energy of the incident photon. $hv = 1/2mv^2 + hv_0 $ By c= f * z where f = frequency and z = wavelength $hc/z = 1/2mv^2 + hv_0$ $z/hc = 1 / (1/2mv^2 + hv_0)$ $z = hc / (1/2mv^2 + hv_0)$


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In order to answer your question you must think through the entire measurement. In this case you are measuring the photoelectric effect by placing a photodiode into a circuit. What kind of circuit element is the photodiode? It is a source. And as you have pointed out above, it is a constant voltage source (with voltage determined by the energy of the photons ...


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You can increase the intensity (Power/Area) in any one of three ways: increasing the rate of incident photons, decreasing the area, or increasing the energy of the photons. (In each case we take the other two as remaining constant.) Increasing the energy per photon will increase the energy of the emitted electrons, but not the rate of emission. ...


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The intensity can only be increased by greater energy of the photon. When you increase the energy of the photon, the energy is absorbed into the electron, which consequently flies off with the photon's energy. The more energy the photon has, the more energy the electron has.



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