# Tag Info

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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

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 ...

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 ...

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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 ...

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

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 ...

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

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 ...

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

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 ...

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 ...

1

The metal's threshold wavelength is a wavelength of light. So yes, you would use a chart converting wavelengths of light to the color to identify it. For some metals, the threshold wavelength is not visible light; it might be ultraviolet. But whatever chart you're using would identify the wavelength you have as either ultraviolet or visible, and which color ...

1

I think you should re-read the question, because "frequency of the metal" doesn't make much sense in this context. And the color of photon is dependant only on it's frequency which is given. Do you need some further explanations on that? Photoeffect's main equation is one of energy balance. The only energy source in this event is the photon whose energy is ...

1

The deduction of the thermal energy distributions are pretty much just Stirling approximation $\ln(x!)=x\ln(x)-x$, Lagrange-multipliers method and a lot of permutations/combinations. You can see it at the bottom. Thermal energy distributions contain classical models such as Maxwell-Boltzmann statistics and quantum-physical models such as Bose-Einstein ...

1

The problem with using a chunk of iron is that in a metal the sharp orbital energies that you get in isolated atoms spread out to form energy bands. Typically when you shine light on an iron surface nothing will happen until the energy exceeds 4.5eV (275nm so that's in the near uv), at which point it will eject photoelectrons. Increasing the energy of the ...

1

At the fundamental level there are four forces associated with the interactions of particles in the microcosm: the strong, the weak, the electromagnetic and the gravitational one. The last two are long range forces and influence the behavior of matter macroscopically too, in a collective manner. Macroscopically phenomena of absorption can be observed which ...

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This is probably related to the derivation of de-Broglie wavelength... Since photon has wave-particle duality, We could equate Planck's quantum theory (wave nature) which gives the expression for energy of a wave of frequency $\nu$, ($E=h\nu$) with Einstein's mass-energy equivalence (particle nature) which gives relativistic energy for photon ($E=mc^2$) ...

1

If it is ionising an atom with a constant field that you want, that is certainly possible in principle and it works in much the same way that ionization in an intense IR field does, via tunnel ionization. The way this works is that the constant field adds a linear potential $V=eE_0 x$ to the atomic Coulomb attraction, which means that in some scale the ...

1

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 ...

1

There are electrons on both plates, collector and emitter. It is true, that there are less negatively charged electrons than positively charged atoms on emitter plate, which makes it positively charged and vice-versa on the collector plate, but there are electrons on both plates. In fact, the total surplus charge on both plates exactly matches the ...

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The first thing you have to calculate is the number of photon incident on the photocathode per unit time! That you can calculate by: $P_{laser}=E_{inc}/dt=[J/sec]$ since every photon carries an energy equal to $h\nu=h\cdot c/\lambda=3.37\cdot 10^{-19} J$ (greater than the work power of course, if not what the heck are we doing here??) you can obtain: ...

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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 $N_{\gamma}= ... 1 The energy of a photon depends on the frequency:$E=h \nu\$. This is not a classical result. So classically: To extract electrons from a material you need a minimal energy. This can be provided by a minimal intensity. But: from a quantum mechanical point of view this reasoning breaks down. That's the discovery of the photoelectric effect (and its correct ...

1

Classically, light was thought to only have the characteristics of wave. The energy of a wave depends solely on the intensity not frequency. To knock out an electron one expects supplying enough energy would make it escape from the nucleon's attraction, so people won't expect frequency affects the photoelectric effect.

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