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29

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, ...


20

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 ...


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 ...


11

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 ...


10

The energy needed to remove an electron from a solid is called the work function. For most metals you would need UV photons (300 nm for Aluminium) that rarely reach the Earth's surface. Visible light can eject electrons from alkali metals, but the quantum yield (the probability of electron emission per incident photon) for pure metals is low (probably less ...


10

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 ...


9

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, ...


9

You can't just get it from the atomic properties, the electronic properties of a metal are dominated by "solid state"-type considerations, for instance, the fact that electrons live in a band structure rather than something more akin to the usual discrete levels that one learns about in QM 1. Thankfully, Ashcroft and Mermin's classic book has a long ...


8

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 calcualtion, 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 ...


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 ...


7

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.


7

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 ...


6

I believe CuriousOne is correct, however it does not make any sense to define the threshold frequency for the photoelectric effect in anything but the rest frame of the metal. From the rest frame of the spaceship, the metal plate is rushing towards it and the apparent threshold frequency is lowered, but an occupant of the spaceship should realise that this ...


6

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 ...


6

There are three main attenuation processes for high energy photons. The photoelectric effect and Compton scattering are more important at lower energies and their cross-section decreases monotonically with increasing energy. On the other hand, the pair production cross-section takes over and increases towards higher energies. The sum of these is what is ...


5

All books say only a single photon can remove an electron at a time That's probably a misstatement. A single photon can remove an electron, but that's not the only way that an electron can escape. The word "only" is the problem. There are two different phenomena going on here. When an object gets warm, the electrons in the material also heat up. You ...


5

The reason is simple. If you make it hemispherical, and shine sunlight vertically on it, less light will reach the photoelectric cells ( due to the formation of shadows )! So, if we put it horizontally, there will be no shadows and most of the light will be used for the desired purpose and in the other case, much of the light is wasted!


5

It is possible and it has already been observed. In the following article they suggest both single and double step emission mechanisms: Two Electron Photoemission in SolidsR. Herrmann, S. Samarin, H. Schwabe, and J. Kirschner Phys. Rev. Lett. 81, 2148 – Published 7 September 1998 The single step emission process is because of correlletions of the two ...


5

The "particle wave duality" comes because the experimental behavior of particles in the microcosm is mathematically described by quantum mechanics. In quantum mechanics a "particle's" position can only be calculated from a probability distribution, the complex conjugate square of the wave function. The wave function is a solution of the quantum mechanical ...


5

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 ...


5

The term Resistance does not come into play while dealing with Photoelectric Effect. The latter is related to the emission of electrons when the surface of a metal(or any substance) is hit by photon particles(photon is the unit particle making up the light that we talk of). Here the more important concept is that of Work Function, i.e. the minimum amount of ...


5

In particle physics there exists elastic scattering for all interactions: change of direction but not of energies. When a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon, the scattering process, also known as Rayleigh scattering, is also elastic. In this scattering process, ...


5

I just remember $$ \frac{1}{\exp(\beta (E-\mu)) \pm 1}$$ You can work out the sign from the fact that Bose-Einstein distributions can diverge (so they go with the - sign), whereas Fermi-Dirac is bounded (so they go with the + sign). Maxwell-Boltzmann applies to classical systems, so quantum statistics don't matter, so take the limit that the two ...


5

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 ...


5

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 ...


4

You obviously don't need a relativistic calculation because the rest mass of an electron is around half a MeV and the energies you're dealing with are orders of magnitude less than this. In fact you got the correct formula: $$ A = \frac{4W_2 - W_1}{3} $$ but you've made a mistake with the numbers somewhere. $W_1$ (350nm) = 3.542eV $W_2$ (540nm) = ...


4

Certainly vinas is correct. The absorbed energy is converted to heat energy. The scenario you mention with the LED is very close to the blackbody problem known as the "ultraviolet catastrophe." There is a Wikipedia article about it here. What happens in the situation you described is that the light proof box gets hotter. It will increase in heat until ...


4

You should definitely use the textbook value of $h$. In any experiment there are always (hopefully) small errors in measurements so the fact that you get two different values for the work function $W$ from two different experiments is to be expected. Average those $W$s and consider their difference to be a rough estimate of the potential error. You say ...


4

Metals do disintegrate in light, just very slowly. Light is often a very weak source and its effect is quite unnoticeable. There are many systems that use highly concentrated light beams, lasers, for etching purposes. http://en.wikipedia.org/wiki/Laser_engraving Note, however, that as electrons are removed, it becomes roughly exponentially harder to ...


4

The work function, $\phi$ is the amount of energy required to free the electron from the pull of the nuclei of the atoms of the photosurface. Here $\phi=3\text{ eV}$. Since the kinetic energy of the electron is given by, $$E_k=hf-\phi$$ it becomes evident that the condition of electron emission is when $hf>\phi$. Clearly this is not the case, which is ...



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