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Unformatted text preview: How does Einsteins photon hypothesis explain all the observational facts? We list here the explanations for the same six observational facts above: 1. Say the intensity is low, i.e., light is very dim: there are few photons irradiating the surface. It is not very likely then that the photon will collide with any atom in the material, and therefore will not eject any electron. If the light is very bright, there are many photons, and a certain fraction thereof collides with atoms and ejects electrons. If we now double the intensity, we also double the number of incident photons, and therefore also double the number of collisions and correspondingly the number of ejected electrons. (Recall that this experimental result is partially explained also by classical theory.) 2. When a photon collides with an atom it ionizes it and releases an electron on the atomic transition time scale. There is no need to build up the energy transferred to the material, and all the energy is imparted at once. 3. For fixed I , if we increase f the number of photons n decreases (for the product nhf to remain constant). With fewer photons, there are also fewer ejected electrons. 4. Below the cutoff frequency the photon does not carry enough energy t release the electron, regardless of how many photons there are. Recall that electrons are released by a single photon. (The probability of multiphoton interaction is very small.) 5. Increasing the intensity means increasing the number of photons, but the energy im parted by a single photons is unchanged. 6. When the frequency f increases, the photon carries more energy. The work function remains unchanged, so that the additional energy is converted to a higher kinetic energy, i.e., higher K max . As we showed above V stop f with the same proportionality coefficients for all materials ( h/e ), and k max = eV stop . The Compton effect If light is made of particles, then photons should collide with other particles like particles do. We therefore analyze the elastic collision of two particles, letting one of them have energy according to Einsteins hypothesis ( E = hf ), and allow for velocities to be relativistic. We take the particle to be static before the collision, and the photon moving before the collision in the x direction. We denote the angle the scattered photon makes with the x axis by and the angle the scattered particle makes with the same axis by . 114 Conservation of energy: hf = hf + (  1) mc 2 where (  1) mc 2 is the recoil relativistic kinetic energy of the particle the photon collides with....
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 Spring '09
 LIORBURKO
 Physics, Photon, Light

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