005626f776RE Posted October 3, 2018 Posted October 3, 2018 (edited) Hello, I am an artist working in the field of photography. I am wondering, is the photoelectric effect considered "physical"? Because I read that photons have no mass... Any explanation or opinion appreciated. I hope I have posted in the right forum. Edited October 3, 2018 by 005626f776RE
StringJunky Posted October 3, 2018 Posted October 3, 2018 (edited) 22 minutes ago, 005626f776RE said: Hello, I am an artist working in the field of photography. I am wondering, is the photoelectric effect considered "physical"? Because I read that photons have no mass... Any explanation or opinion appreciated. I hope I have posted in the right forum. If something can be measured it is physical; it doesn't have to have mass to qualify. To put it another way: if something can be related to physics, it's physical. Photons are physical. The notion of 'physical' in the scale of our everyday experience is related to things we can sense through touch and have some kind of solidity but when you zoom in far enough everything is just fields and, at that small scale, the idea we call "solid" is actually just repulsion between sets of these fields, like magnets, which actually have no sense of substance we can directly relate to. Edited October 3, 2018 by StringJunky 1
005626f776RE Posted October 3, 2018 Author Posted October 3, 2018 Thanks so much for the response! May I ask?: How does something without mass can cause perturbation to a photosensitive surface? Is this related to energy? And can we have energy without mass? many thanks
Sensei Posted October 3, 2018 Posted October 3, 2018 14 minutes ago, 005626f776RE said: Because I read that photons have no mass... In Standard Model, having "no mass" often means having "no rest-mass"... Can you put photon to rest? No.. Particles with rest-mass can be accelerated on demand (in lab), decelerated on demand.. they can have various different speeds.. Photoelectric effect can be observed by naked eye, without any device. In XIX century, when photoelectric effect has been discovered, there was no precise electronic/electric instruments. 1 minute ago, 005626f776RE said: Thanks so much for the response! May I ask?: How does something without mass can cause perturbation to a photosensitive surface? Is this related to energy? And can we have energy without mass? many thanks Life on the Earth exists just because of light (=photons) from the Sun... Photons are absorbed/scattered, and increases energy of atoms they hit. Atoms are excited. Which can be detected by humans as e.g. increase of temperature. New photons (typically) with lower energies are emitted (in IR/MW spectrum range). Matter absorbs energy, and releases energy, all the time. 1
StringJunky Posted October 3, 2018 Posted October 3, 2018 (edited) 16 minutes ago, 005626f776RE said: Thanks so much for the response! May I ask?: How does something without mass can cause perturbation to a photosensitive surface? Is this related to energy? And can we have energy without mass? many thanks Photons have energy and they can be absorbed by an electron, if it has the right frequency. This absorption causes the electron to become energetic and that can create a current in a digital system, or cause a film emulsion to deposit the metal, which makes your negative or slide image in one spot of it. Edited October 3, 2018 by StringJunky 1
005626f776RE Posted October 3, 2018 Author Posted October 3, 2018 Many thanks! One last question for now, is the photon that is reflected from a photographed surface the same as the one that "hits" my photosensitive surface (camera) causing a photoelectric effect?
StringJunky Posted October 3, 2018 Posted October 3, 2018 (edited) 7 minutes ago, 005626f776RE said: Many thanks! One last question for now, is the photon that is reflected from a photographed surface the same as the one that "hits" my photosensitive surface (camera) causing a photoelectric effect? Can you rephrase that? There's no photoelectric effect when you view an image; The photon is absorbed and immediately reflected back at you, or the ones incident to your eyes. Edited October 3, 2018 by StringJunky 1
005626f776RE Posted October 3, 2018 Author Posted October 3, 2018 (edited) When I photograph a stone, does the photon that hit the stone get reflected and then hits my photosensitive surface in the camera? Is this the same photon? Edited October 3, 2018 by 005626f776RE
Sensei Posted October 3, 2018 Posted October 3, 2018 2 minutes ago, 005626f776RE said: Many thanks! One last question for now, is the photon that is reflected from a photographed surface the same as the one that "hits" my photosensitive surface (camera) causing a photoelectric effect? Photoelectric effect is different from photosensitivity... https://en.wikipedia.org/wiki/Silver_halide In photoelectric effect photon is absorbed, and slower/faster electrons are ejected.. 1
StringJunky Posted October 3, 2018 Posted October 3, 2018 (edited) 5 minutes ago, 005626f776RE said: When I photograph a stone, does the photon that hit the stone get reflected and then hits my photosensitive surface in the camera? Is this the same photon? Essentially yes, the reflected photon that enters your lens has the same properties as the photon that hit the stone. Edited October 3, 2018 by StringJunky 1
005626f776RE Posted October 3, 2018 Author Posted October 3, 2018 (edited) Quote Essentially yes, the reflected photon has the same properties as the photon that hit the stone. So is it the exact same particle? or just identical? Edited October 3, 2018 by 005626f776RE
Sensei Posted October 3, 2018 Posted October 3, 2018 Just now, 005626f776RE said: So is it the exact same particle? or just identical? "The same particle" is indistinguishable from "particle with the same properties".. 1
StringJunky Posted October 3, 2018 Posted October 3, 2018 (edited) 11 minutes ago, 005626f776RE said: So is it the exact same particle? or just identical? I asked this once and was asked "Does it matter?". Two photons with the same frequency are the same. The electron absorbs the photon, and because it's the 'wrong' energy, emits it back out again in exactly the same form. If it was the 'right' energy, it would be absorbed and you would see black. There will be no reflection. Edited October 3, 2018 by StringJunky 1
swansont Posted October 4, 2018 Posted October 4, 2018 15 hours ago, Sensei said: Photoelectric effect can be observed by naked eye, without any device. That's deceptive, at best. You can observe a spark with the naked eye, but "without any device" does not mean you can do this with without equipment. 15 hours ago, Sensei said: In XIX century, when photoelectric effect has been discovered, there was no precise electronic/electric instruments. They were precise enough to measure voltages and/or currents, which is what you need. 15 hours ago, 005626f776RE said: Thanks so much for the response! May I ask?: How does something without mass can cause perturbation to a photosensitive surface? Is this related to energy? And can we have energy without mass? many thanks Energy, and also momentum. Light being absorbed will push on the object that absorbs it, as will emitting the light, similar to a skater catching a ball or throwing it; they will feel a recoil from each. 1
005626f776RE Posted October 4, 2018 Author Posted October 4, 2018 (edited) Thank you all for the responses so far. Can we say then that photography is possible solely because of the photoelectric effect, the latter affecting equally analog photosensitive surfaces and digital photosensitive sensors? Edited October 4, 2018 by 005626f776RE
swansont Posted October 5, 2018 Posted October 5, 2018 14 hours ago, 005626f776RE said: Thank you all for the responses so far. Can we say then that photography is possible solely because of the photoelectric effect, the latter affecting equally analog photosensitive surfaces and digital photosensitive sensors? No. Photography, either film or digital, does not rely on the PEE. Red light does not have enough energy to ionize (which is the result of the PEE), so you couldn't capture red, or infrared, in a system that relied on it. Both film and digital rely on excitation of electrons, which is possible with lower energy; in film this facilitates a chemical reaction. In digital the electrons are stored and then counted. 1
005626f776RE Posted October 5, 2018 Author Posted October 5, 2018 (edited) Thanks for the explanation! How is this event triggered exactly? Does not PEE precede it?: Quote Both film and digital rely on excitation of electrons I just read this on Wikipedia "Electron excitation is the transfer of a bound electron to a more energetic, but still bound state. This can be done by photoexcitation (PE), where the electron absorbs a photon and gains all its energy" How is PE different from PEE? What is the sequence of events from photons reflected from a photographed surface(for example a stone) to the chemical reaction/ digital storage? many thanks Edited October 5, 2018 by 005626f776RE
Sensei Posted October 5, 2018 Posted October 5, 2018 @005626f776RE Photoelectric effect is creating free electrons. Photon is completely absorbed and disappears from the system. 1
005626f776RE Posted October 5, 2018 Author Posted October 5, 2018 Quote Photon is completely absorbed and disappears from the system. nice explanation. It helps a lot. Could you please help me with the sequence of events below? photon hits electron ----> electron absorbs photon ---> then up to the digital storage/chemical reaction?
swansont Posted October 5, 2018 Posted October 5, 2018 43 minutes ago, Sensei said: @005626f776RE Photoelectric effect is creating free electrons. Photon is completely absorbed and disappears from the system. That's true in either case. The photon is absorbed and disappears. What's different is whether the electron is free or still bound. In the PEE, the electron is free. With excitation, it is still bound. 32 minutes ago, 005626f776RE said: Could you please help me with the sequence of events below? photon hits electron ----> electron absorbs photon ---> then up to the digital storage/chemical reaction? Technically you can't say what is absorbing the photon (it's the whole atom), but yes, basically. A photon is absorbed and an electron is promoted to a higher energy state, and either remains there until measured (digital), or facilitates a chemical reaction (film). 1
005626f776RE Posted October 5, 2018 Author Posted October 5, 2018 I have a huge appreciation for your answers. Based on these quotes: Quote an electron is promoted to a higher energy state Quote that can create a current in a digital system How exactly is the electron with higher energy physically affecting/continuing in the digital system? many thanks
swansont Posted October 5, 2018 Posted October 5, 2018 1 hour ago, 005626f776RE said: How exactly is the electron with higher energy physically affecting/continuing in the digital system? many thanks https://en.wikipedia.org/wiki/Charge-coupled_device 1
005626f776RE Posted October 6, 2018 Author Posted October 6, 2018 (edited) many thanks! I already studied the CCD from wikipedia and youtube Edited October 6, 2018 by 005626f776RE
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now