# Tag Info

24

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 ...

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, ...

6

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 ...

6

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 ...

6

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 ...

6

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 ...

5

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 ...

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 ...

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

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 ...

4

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, ...

4

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 ...

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

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

In a small metal bar (let's say 1cm of length, a few grams of weight) there are about $10^{24}$ protons, amounting to a total charge of $10^{14}$ e.s.u. The interaction energy between these protons is $Q^2/r$, which amounts to $10^{28}$ erg, or $10^{21}$ J. This is repulsive energy, of course, so that the value we just estimated is the work we would need to ...

3

What you are describing might be possible - that a series of interactions with photons might cause an electron to first be excited, and then to be emitted. However, the probability of an individual electron interacting with two photons in quick succession without an intermediate interaction with other electrons (and thus an opportunity to shed the excess ...

3

The configuration that you describe (all electrons leaving the metal) would be impossible to generate by using a light source. The electrostatic repulsion of two point charges is given by Coulomb's law: $F=\frac{k_e q_1 q_2}{r^2}$ If the positive (protons) and negative (electrons) charges occur in roughly the same quantity, then the material is ...

3

You're quite correct that the plate will get a positive charge, but then any piece of metal can be positively or negatively charged in many ways. For example a Van de Graaff generator does exactly this, as does a gold leaf electroscope. The charges involved are tiny compared to the number of mobile electrons in the metal and make no significant difference to ...

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

it usually gets converted to heat... unless it bounces off or goes right through. That's why black things get hotter than white things. I am not sure about this.

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

Describe the process at a high level: The laser produces a certain number of photons per second $N_{\gamma}$ The photons free a certain number of electrons per second $N_e$ The free electrons per second give you the current $I$ Now $N_{\gamma}= \frac{P_{laser}}{E_{\gamma}}$, where $E_{\gamma} = h\nu = \frac{hc}{\lambda}$ Therefore $N_{\gamma}= ... 3 Well, it will become lighter with each electron removed, but unless you do it in the vacuum, it will get its electrons back from the environment. Also, unless we are talking about really high energies, only the weakly bound valence electrons will be removed. The work to remove the electron,$W_a$, becomes higher the more electrons have left. Okay, now if ... 3 Energy exchange is quantized when moving a electron from one bound state to another bound state. This isn't because the exchange is inherently quantized, but because the states the electron may occupy are quantized. Thus the standard photo-electric effect in which a photon can not excite an atom unless it has a minimum energy. However,... There are ... 3 This is the analogue of a projectile getting launched at exactly the escape velocity, something you may remember from studying gravity in freshman physics. Here we're talking about the photoelectric effect. The electron jumps out of the material into air or vacuum, overcoming the force of attraction that tries to keep it bound inside the material (the force ... 3 First, note that the metal "plate" will need to be more like a strip -- it needs to be small enough to fit into one trough (or peak) in the interference pattern. But this thin strip that interacts with light is just like any other photodetector (including your eye). So, when it is in a trough, no electrons are emitted. Your intuitive picture of "two ... 3 The work function,$\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 ...

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

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 ...

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