40

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 ...


33

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 ...


32

The Lamb-Scully paper is a good example of how even a Nobel Prize winner can occasionally write a bad paper. The historical context is important. Einstein hypothesized the photon in 1905, but his paper was ahead of its time and was not widely accepted. For decades afterward, even once the quantum-mechanical nature of the atom was assumed by all physicists, ...


23

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 ...


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 ...


20

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 ...


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 "...


18

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 ...


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 ...


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

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.


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 ...


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

For a given system that the electron is in, the primary determinant is the energy of the photon. As @DJBunk points out, this is a quantum mechanical process, so the "choice" is fundamentally random. A given interaction will occur with a probability proportional to its cross section. Figure 1 of this lecture shows how the cross section for each possible ...


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

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 ...


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 ...


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

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 (...


8

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 ...


8

A light ray has an oscillating electric field associated with it, and this oscillating electric field will make electrons oscillate when the light ray passes through them. Note that I'm talking about light rays not photons - we'll get to photons in a bit. When the oscillating field of the light makes electrons oscillate energy can be transferred between the ...


7

Why, after absorbing a photon does an atom's electron 'fall' back to its ground state (what causes it to immediately lose its absorbed energy)? The answer by @Davidmh gives our observations from classical physics, where we formulated the quantities "energy" , "potential" etc. We observed that this was so, an apple falls, and brilliant mathematics organized ...


7

But the stopping potential does depend on the kinetic energy of the electrons. The stopping potential is defined as the potential necessary to stop any electron (or, in other words, to stop even the electron with the most kinetic energy) from 'reaching the other side'. As you already stated, the maximum kinetic energy is given by $$K_\text{max}=h\nu-e\phi_\...


7

There is a recoil when each photon leaves, but they radiate in all directions at once, as ACuriousMind intimates in his comment, so there is no collimated beam to concentrate the recoil. Even if the total recoil were concentrated, its effect is so small that the screw-in base of the bulb is more than sufficient to hold the bulb steady. Theoretically, it ...


6

It may be a reference to the fact that you can reproduce the characteristics of the photoelectron production in a model which treats the incident light classically, but treats the matter in the target quantum mechanically. This is explained in Mandel and Wolf's book (chapter 9), which explains how a simple semiclassical calculation can be used to derive the ...


6

I just had an exam today with a similar question. I guess you think of light intensity as amount of power/sec hitting the metal. Thus if we increase the intensity (but keep the frequency of light constant), all we are doing is adding more photons with the same amount of energy. And the electrons by rule can only absorb all or none of the energy provided by a ...


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