# Tag Info

## Hot answers tagged photoelectric-effect

21

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, ...

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 ...

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

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 ...

5

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 ...

5

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 ...

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

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) = ...

3

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 ...

3

This is the analogue of a projectile getting launched at exactly the escape velocity, something you may remember from studying gravity in freshman physics. Here we're talking about the photoelectric effect. The electron jumps out of the material into air or vacuum, overcoming the force of attraction that tries to keep it bound inside the material (the force ...

3

It's common to think of the light as photons that behave like billiard balls colliding with and knocking out electrons. However this isn't a good model for what actually happens. The photon is really the quantised energy transfer from the photon field to the metal. Localisation of the light into a photon only happens at the moment the electron is ejected. ...

3

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 ...

3

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 ...

3

Energy exchange is quantized when moving a electron from one bound state to another bound state. This isn't because the exchange is inherently quantized, but because the states the electron may occupy are quantized. Thus the standard photo-electric effect in which a photon can not excite an atom unless it has a minimum energy. However,... There are ...

3

Well, it will become lighter with each electron removed, but unless you do it in the vacuum, it will get its electrons back from the environment. Also, unless we are talking about really high energies, only the weakly bound valence electrons will be removed. The work to remove the electron, $W_a$, becomes higher the more electrons have left. Okay, now if ...

3

First, note that the metal "plate" will need to be more like a strip -- it needs to be small enough to fit into one trough (or peak) in the interference pattern. But this thin strip that interacts with light is just like any other photodetector (including your eye). So, when it is in a trough, no electrons are emitted. Your intuitive picture of "two ...

3

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, ...

3

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 ...

2

You're mistaken in your history. The number of electrons emitted does, in fact, increase with the intensity of the radiation. Higher intensity means more photons, which means more chances to knock an electron loose. It's the energy of the emitted electrons that people expected to increase that doesn't. In a classical wave picture, an increased intensity ...

2

Using that ideal white source on the wiki link, that states 251 lm/W. The insolation level is $2.61kWh/m^2d = \frac{2610}{24}W/m^2$ which across $65cm^2 (= 0.0065m^2)$ gives 0.7W. If everything were 100% efficient, then you'd have $0.7 \times 251 lm = 178 lm$ Now we derate on some maximum theoretical efficiencies. The biggest theoretical derating will be ...

2

The 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 ...

2

Note the power units given for the laser intensity. Power is energy transferred or transformed per unit time. So, how many photons per second, and from that how many photo-electrons? And an interesting question to ask yourself then is if the metal is electrically isolated, can this go one forever, and if not why and how does it stop? What would you need ...

2

In the photoelectric effect one photon displaces one electron, so the way to understand it is to consider the properties of the photons. If you take light of a fixed frequency, $\nu$, then the energy of the photons is fixed at $h\nu$. That means the intensity of light is proportional to the number of photons, and because one photon = one displaced electron, ...

2

The photon is the particle that carries the electromagnetic force i.e. charges exert a force on each other by exchanging virtual photons. In your example of a capacitor one plate has a positive charge and the other has a negative charge, and the two plates are continuously exchanging virtual photons, which causes the attractive force between the two plates. ...

2

The "color" of a photon can be ultraviolet. Visible light is just a small part of the electromagnetic spectrum. Ultraviolet is the part of the spectrum with slightly shorter wavelengths than blue and purple. Many materials have a threshold wavelength in the ultraviolet. And for any material with a threshold wavelength in the visible, ultraviolet light will ...

2

Edges are the vertical lines in the graph. As you increase the photon energy, it suddenly becomes possible to excite an electron from the interior of the atom – from an occupied state – and move it to an unoccupied state i.e. to ionize an atom. The K,L,M,N... shells are names for the states corresponding to the principal quantum numbers $n=1,2,3,4,\dots$ ...

2

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 ...

2

Suppose the potential at the surface of the ball is $+V$, then the work required to remove an electron from the surface to infinity is simply $+V$ eV. The kinetic energy of the electrons is (using your notation) $W_{k0}$ eV, so the photoelectrons will be unable to escape the ball when $V = W_{k0}$. All you have to do is calculate the voltage of the sphere as ...

2

In particle interactions the total number of particles is not conserved. For example in a collision in the LHC two photons collide and many hundreds of particles are created in the collision. There are still some conserved quantities, for example lepton number is still conserved so you cannot just create an electron. You need to create an electron and ...

1

I would guess the question is asking you what the maximum charge on the sphere is. Suppose the photon energy ($hc/\lambda$) is $E$, then the kinetic energy of the electron leaving the surface (in electron volts) is $E$ - 4.47. As you increase the positive charge on the sphere you increase the work needed to remove an electron to infinity, and for some ...

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