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Jun 3, 2015 at 15:42 comment added TheQuantumMan So you see that it was not a silly question ;) thank you
Jun 3, 2015 at 15:38 comment added WhatRoughBeast "It is possible, but usually happens seldom enough that it can be ignored unless the intensity of the beam is extremely high." See curious.astro.cornell.edu/physics/137-physics/general-physics/…
Jun 3, 2015 at 14:38 comment added TheQuantumMan I do not want a metaphysics explanation.Your answer just explains me what we get from experimental results(the graph),but in physics you have to explain every detail.So what i am asking is,why do we say that one electron absorbs maximum one photon?Why not two for example?
Jun 3, 2015 at 14:34 comment added WhatRoughBeast I'm sorry, but you are asking WHY in a manner which I cannot understand. Given an experimental result, I (or in this case, Einstein) gave a compelling explanation. But you ignore/dismiss the explanation and ask WHY? Sorry, but this is a physics site, not a metaphysics site.
Jun 3, 2015 at 14:29 comment added TheQuantumMan But what is the explanation of this?I mean you are explaining to me the expirimental observation(the curve).But WHY is this happening?
Jun 3, 2015 at 14:27 comment added WhatRoughBeast Because the photocurrent vs voltage curve is smooth. If not, you'd get most of the curve occurring with one photon per electron, then at high voltages you'd get a change in the curve as you start dealing with the dual-photon electrons.
Jun 3, 2015 at 14:23 comment added TheQuantumMan So,why do we make the assumption that only ONE photon transfers energy to an electron?
May 31, 2015 at 11:20 comment added WhatRoughBeast And "when we have light of higher frequency then we have more photons giving energy" is not right. For a given brightness, higher frequency means fewer photons. Brightness is energy per unit time. If each photon has certain amount of energy, increasing the individual energy while keeping the total energy (brightness) the same means fewer photons per second.
May 31, 2015 at 11:18 comment added WhatRoughBeast Nope. That's the point. The photoelectric effect does not behave like a battery and a resistor. Brightness (intensity) is just what it sounds like. It's measured in power per unit area. Shine your light through an aperture and you get a certain amount of power (energy per second) on the target. Cut the aperture size in half, and you get half the brightness. If the frequency is above threshold, you get half the current. If the frequency is below, no current regardless of brightness...
May 31, 2015 at 8:47 comment added TheQuantumMan i mean,if the energy from the photons is like the energy given by a battery in a circuit,then the energy of the photons is the one that determines the current.I do not understand what the intensity-brightness changes.Can you explain to me by giving an analogy to a circuit with a battery?Thank you
May 31, 2015 at 8:45 comment added TheQuantumMan Can i ask something else just to clarify the things that you say?Well,i have a question about the current and the energy.If we have a battery in a circuit,the higher its potential the greater the current.So in this situation,when we have light of higher frequency then we have more photons giving energy to electrons at a given time(the chunks that i mentioned) and then if the frequency is high enough the electrons get kicked loose.So there is the energy.I do not understand what the intensity(brightness) has to do with the current.
May 31, 2015 at 0:37 history edited WhatRoughBeast CC BY-SA 3.0
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May 31, 2015 at 0:31 history answered WhatRoughBeast CC BY-SA 3.0