44

Yes,the photoelectric effect can be explained without photons! One can read it in L. Mandel and E. Wolf, Optical Coherence and Quantum Optics, Cambridge University Press, 1995, a standard reference for quantum optics. Sections 9.1-9.5 show that the electron field responds to a classical external electromagnetic radiation field by emitting ...


32

This classical prediction comes from the equipartition theorem of statistical mechanics, though I have some issues with exactly how the statement you quote is worded. The equipartition theorem is for describing how the energy gets distributed in a system with many degrees of freedom. For example, consider a mono-atomic ideal gas, like helium, that you've ...


29

This is an answer by a particle physicist that has been working with data for forty years: Photons and electrons are quantum mechanical entities, and to really really understand their interactions, quantum mechanics has to be invoked. When detected, the photon has a point particle footprint (as does the electron) consistent with the axiomatic particle ...


27

For a photon to give rise to a real (not virtual) electron/positron pair it must possess an energy of slightly greater than one million electron volts. This is a very energetic photon indeed. In comparison, the photon causing photoejection of an electron from an atom needs an energy of order ~an electron volt. This is typical of the photons that make up ...


26

Rather than considering quantum efficiencies or such details it's instructive to step back and take a broader view. One of the main fuel crops grown in the UK is miscanthus. There are various figures around for the yield produced by miscanthus, but these people estimate it as about 14 tonnes per hectare per year. The energy content is 19GJ/tonne, so that's ...


23

It depends what you mean by "size". Light spreads out like a ripple in water so once that ripple reaches some object floating in the water it will disturb the object. Does the ripple have some definable "size"? It's just an ever expanding circle whose source is the center of the expanding circular ripple (maybe caused by you dabbing your finger in the water ...


22

For simplicity let's consider the photoelectric effect in a thin metal foil: The first step in the photoelectric effect is when a photon strikes an electron in the metal and transfers all its energy to it. The electron energy is now equal to the photon energy $h\nu$. If this energy is greater then the work function $\phi$ the electron can escape the metal ...


22

The metals that are used in photoelectric experiments belong to the first group of the periodic table. They are often called Alkali metals. They have the highest electropositive nature in their respective periods. This makes them the most reactive. Any reactive metal would not exist in nature in its elemental state. So, these metals get oxidised by oxygen ...


19

In general you're right - an electron being subject to interactions with more than a single photon may have a higher kinetic energy. However, in the vast majority of photoelectric setups you will observe that kinetic energy is independent of light's intensity. The appropriate framework for this discussion is this of probability theory: Each electron has an ...


19

Sometimes the effects do become visible in electronics. One example is the case where Raspberry Pi 2 could be crashed by camera flashlight: Upton explained that the semiconductor material used to make the power regulator was subject to a photoelectric effect when hit with light, and if enough light of the right energy was fired at it, then it would "...


17

Yes, the textbooks are getting it very wrong. The common narrative on these things is best summarized by the "three nails in the coffin" approach: the dead body being the wave theory of light, and the three nails being the blackbody spectrum, the photo-electric effect, and the Compton effect. Whatever difficulties the wave theory may or may not have with ...


17

The answer is, it does happen. Just at largely different energies. This picture (taken from this thesis, page 10) summarizes it quite nicely for scattering on Cu (Copper) atoms: Photoelectric absorption is just the most relevant effect at low energies (as you can see from $10 - 10^3$ eV. Let me summarize the other effects ($A$ stands for atom): Thomson ...


16

Photosynthesis is less efficient than solar panels. According to the Wikipedia page on photosynthetic efficiency, typical plants have a radiant energy to chemical energy conversion efficiency between 0.1% and 2%. Most commercially available solar panels have more than 10 times this efficiency.


16

The problem is that you are confusing light intensity with energy of a single photon. The photoelectric effect requires a certain energy per photon to work. But low light intensity just means fewer photons come - you can actually see the grain if the conditions are too dark: every pixel can get ~10 photons or less... and yet still, each photon that comes has ...


15

This is actually a good question. Because of conservation of momentum/energy, this cannot happen. A single photon will not decay into an electron-positron pair such that they would exist as two separate and independent particle states. One of the particles has to be "off-shell". In other words the process of pair-production will involve a virtual ...


14

They do, but it's too small to notice on a human scale. On the scale of electronics, you absolutely can see it. We have photoresistors and photodiodes which rely on this effect. You need to be measuring this with a multimeter and looking at changes of resistance though - it's far too small for it to be perceptible as a static shock. For another use which ...


13

More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$. The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling ...


12

It is somewhat matter of what precisely one would refer to as photoelectric effect. As far as the radiation-electron mechanism of transfer of energy, there is no direct role played by surface. However, referring to the Einsten's formula; $$ h f = \Phi + K, $$ where $K$ is the maximum kinetic energy of the photoelectron, $f$ the frequency of the incoming ...


11

I disagree with OP in that I don't consider energy conservation as a fatal flaw. If one lets $t\to\infty$ in the perturbative calculation, one gets a nice delta function $\delta(\epsilon_f-\epsilon_i-\hbar\omega)$ but in such case the external energy supply is infinite and no meaningful energy conservation argument can be formulated, so I guess OP must be ...


11

There is so called equipartition theorem in classical statistical physics that says that under some conditions all degrees of freedom have the same average energy if the temperature is fixed. The degrees of freedom of electromagnetic field satisfy those conditions, and there is an infinite number of such degrees of freedom (frequencies can be arbitrarily ...


10

does this mean that Ohm's law just fails in this case Ohm's law is not universal. The ideal resistor circuit element is defined by Ohm's law but not all circuit elements obey Ohm's law; Ohm's law only applies to ohmic devices. Physical resistors and conductors approximately obey Ohm's law but, for example, semiconductor diodes, transistors, thyristors, ...


10

In the photoelectric effect, photons incident on the cathode cause the emission of electrons. Assuming there is a sufficient electric field, these electrons will make their way across to the anode, contributing current. For simplicity, let's assume every photon generates a photo-electron. Then if $N$ photons per second hit the cathode, the current will be ...


10

What Einstein added to the discussion was the idea that electromagnetic energy comes in little particle-like packets. That was a very radical concept at the time (and frankly still is).


10

This is in the x-ray region and beyond. The wavelength of the light is smaller than the size of the electron orbitals, and decreasing when the photon energy goes up. When the electric field oscillates a lot on the length scale of the wave function, positive and negative contributions to the integrals in the transition to the excited state (the ...


10

Both photons and electrons may be considered point-like particles, but the interaction/force that they feel has a range: the electromagnetic interaction has a pretty long range. Actually it is infinite in the absence of screening effects (ideal cases). You could ask yourself, what does it even mean colliding? For example when you clap your hands, the atoms ...


9

Yes excited states have a non-zero lifetime. Electronically excited states of atoms have lifetimes of a few nanoseconds, though the lifetime of other excited states can be as long as 10 million years. The decay probability can be calculated using Fermi's golden rule. The lifetime is then an average lifetime derived from the decay probability. The lifetime ...


9

In the context of the photoelectric effect, the key thing to remember is that you get one electron per photon. Current is a measure of how much charge flows per unit time, which is proportional in this case to the number of electrons per unit time, and therefore the number of photons per unit time. The energy of the electrons doesn't come into it at all. ...


9

Photoelectric effect can be observed instantaneously when light is flashed and it is independent of intensity of the light. If wave model were at play here, intensity of the light (amplitude of the wave) would have an effect on how quickly an electron receives sufficient energy to be knocked out. This is because energy in a wave is related to its amplitude.


9

The energy is delivered to the metal in discrete packets. If you think of light as a wave, than one may expect that a low-energy colour of light should be able to (eventually) liberate electrons from the metal if you wait long enough (as more and more energy is deposited into the metal). Because electrons are not liberated below a certain energy threshold (...


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