Photoelectric effect vs Compton scattering

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  • #26
sophiecentaur
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But this is many-body physics! If you think you can, for example, derive the Fermi Liquid Theory via adding up such single interactions, then I want to see it.

The definition of a conductor here is that there is no gap at the Fermi level! Period!

Now, if you wish to argue that you can show that there is a "just finite" gap here, then I want an exact reference. It is that simple.

Zz.
Would you also say that "main body Physics" insists that a gas is a continuum because it uses the gas laws? Those gas laws came from the statistics of large numbers of discrete particles and we all live with that.
They are only models, remember.
 
  • #27
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I would say that it is misleading to introduce 'work function' as part of the explanation of electrical conduction in metals.
 
  • #28
ZapperZ
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Would you also say that "main body Physics" insists that a gas is a continuum because it uses the gas laws? Those gas laws came from the statistics of large numbers of discrete particles and we all live with that.
They are only models, remember.
Then derive using your model the DOS for free electron gas and show that it has a "just finite gap" at the Fermi energy. So far, you have made claims with ZERO materials to support them.

Zz.
 
  • #29
sophiecentaur
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Why are you being so precious about this? The support for my statement is that calculus is used for that model of 'yours'.
Btw, do you know what the word 'calculus' means?
 
  • #30
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I would say that it is misleading to introduce 'work function' as part of the explanation of electrical conduction in metals.
I would say it isn't, since we're discussing photoelectric effect. Do you know Einstein's formula for the photoeffect?
 
  • #31
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Yes I do and that equation describes the process of ejecting electrons from a metal, not conduction of electricity through a metal.
I wonder if SC is confusing the small ( but not zero) energy gap between donor or acceptor impurities and the conduction or valence band in semiconductors with the continuous conduction band in metals ?
 
  • #32
sophiecentaur
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There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?
 
  • #33
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There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?
Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.

While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands exactly.

The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.

Let us estimate the difference between subsequent energy levels. For this, assume a crysta lsystem with a cubic unit cell with side a. The total number of atoms in the system is N, so that [itex]N = (L/a)^3[/itex]. The discrete values for the wave vector are:
[tex]
k_{i} = \frac{2 \pi \, n_i}{L}, \ (i = x, y, z)
[/tex]
Assume the free electron model so that the energy eigenvalues are:
[tex]
E = \frac{\hbar^2 \, k^2}{2 m}
[/tex]

Then, the uncertainty in energy due to a neighboring level is:
[tex]
\Delta E = \frac{\hbar^2 \, \vert k_i \vert}{2 m} \, \Delta k_i
[/tex]
The order of magnitude for the wave vector corresponds to the Fermi wavevector, found from the total electron density n:
[tex]
\frac{N_e}{a^3} = 2 \frac{1}{(2\pi)^3} \, \frac{4 \pi k^{3}_{F}}{3} = \frac{k^{3}_{F}}{3 \pi^2} \Rightarrow k_F = \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}}
[/tex]
where [itex]N_e[/itex] is the number of electrons per atom.

The uncertainty in the wave vector is given by:
[tex]
\Delta k_i = \frac{2 \pi}{L} = \frac{2 \pi}{a} \, N^{-\frac{1}{3}}
[/tex]

Then, we have:
[tex]
\Delta E = \frac{\hbar^2}{2 m} \, \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}} \, \frac{2 \pi}{a} \, N^{-\frac{1}{3}}
[/tex]
[tex]
\Delta E = \frac{1}{4} \, (\frac{3}{\pi})^{1/3} \, \frac{h^2}{m \, a^2} \, \left( \frac{N_e}{N} \right)^{\frac{1}{3}}
[/tex]

For typical crystals [itex]a \sim 5 \stackrel{o}{A}[/itex]. Take an Avogadro number of atoms, and one electron per atom ([itex]N_e = 1[/itex]). We have:
[tex]
\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}
[/tex]
 
  • #34
sophiecentaur
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Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.

While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands exactly.

The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.
Thanks for that. I take it that this bit is the 'nub' of what you're saying. That makes excellent sense, the fact being that you only need to have your system in an environment to be 'stirring things up' enough to blur out any discreteness of levels that a simple model would suggest.
I was thinking that the 'line spreading' effect as you increase the pressure in a gas (due to the additional interactions) was all that counts but the system wouldn't, even then, be isolated.

Well, at least that's something!. In a real situation, there would be quite a small upper limit to the number of atoms for what I suggested to be actually true. You have accounted for the 'gear change'.
 
  • #35
vela
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I understand that in photoelectric effect, the energy of the whole photon is absorbed, freeing an electron. I don't understand why in the case of Compton scattering, the higher energy photon lost part of its energy instead of transferring the whole of its energy to the electron as in photoelectric effect? Does it mean that there is a limit that the electron could absorb so much energy and no more?
Thanks.
If you analyze the absorption of a photon by a free electron, you'll find you can't conserve both energy and momentum, so with Compton scattering, the scattered photon has to be there. With the photoelectric effect, the rest of the atom absorbs the extra momentum, so the electron can effectively take off with all the energy provided by the photon.
 
  • #36
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If you analyze the absorption of a photon by a free electron, you'll find you can't conserve both energy and momentum, so with Compton scattering, the scattered photon has to be there. With the photoelectric effect, the rest of the atom absorbs the extra momentum, so the electron can effectively take off with all the energy provided by the photon.
First I have to admit that I am not very familiar with the Compton scattering. I just wonder if I assume that in an "absorption of a photon like" interaction of the photon and electron, the whole photon momentum is passed to the electron and as a result the electron gets the relativistic kinetic energy (γ-1)m0c^2 instead of 1/2mv^2
Would that solve the problem (that part of the photon has to remain)?
 
  • #37
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'the rest of the atom absorbs the extra momentum, so the electron can effectively take off with all the energy provided by the photon'.
The ejected electron has a maximum KE of (hf - W) where hf is the energy of the photon and W is the work function of the metal
 
  • #38
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For typical crystals [itex]a \sim 5 \stackrel{o}{A}[/itex]. Take an Avogadro number of atoms, and one electron per atom ([itex]N_e = 1[/itex]). We have:
[tex]
\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}
[/tex]
Dickfore, thanks for that, I’m always grateful for an worked out example. But I’m afraid you only proved SC’s argument to be correct. If I understand you correctly then there would be an extra energy gap of nearly 30 μeV when adding 1 atom to a total of 6E23 atoms. That energy gap is much bigger then I imagined. I can nearly measure such a voltage on my poxy £8.59 multimeter. No need for lab equipment capable of measuring 10^-12 eV and smaller.
 
  • #39
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Dickfore, thanks for that, I’m always grateful for an worked out example. But I’m afraid you only proved SC’s argument to be correct. If I understand you correctly then there would be an extra energy gap of nearly 30 μeV when adding 1 atom to a total of 6E23 atoms. That energy gap is much bigger then I imagined. I can nearly measure such a voltage on my poxy £8.59 multimeter. No need for lab equipment capable of measuring 10^-12 eV and smaller.
The Boltzmann constant is 8.6e-5 eV/K.

You might stand a chance of measuring this below a temperature of 0.3K. I agree that the price of the voltmeter is completely negligible...
 
  • #40
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The Boltzmann constant is 8.6e-5 eV/K.

You might stand a chance of measuring this below a temperature of 0.3K. I agree that the price of the voltmeter is completely negligible...
Nah…. Not in my UK kitchen, all should work out fine.
 
  • #41
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If I understand you correctly then there would be an extra energy gap of nearly 30 μeV when adding 1 atom to a total of 6E23 atoms.
We're talking about electrons here, and the energy separation of their energy eigenvalues in a solid composed of 1 mole of atoms, each contributing 1 valence electron.
 
  • #42
vela
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First I have to admit that I am not very familiar with the Compton scattering. I just wonder if I assume that in an "absorption of a photon like" interaction of the photon and electron, the whole photon momentum is passed to the electron and as a result the electron gets the relativistic kinetic energy (γ-1)m0c^2 instead of 1/2mv^2
Would that solve the problem (that part of the photon has to remain)?
If you assume the free electron absorbs the photon, you'll find the assumption leads to contradictions. If you conserve momentum, you'll find energy isn't conserved, and vice versa. So the logical conclusion is that the electron can't absorb a photon. Only with a scattered photon in the picture can you conserve both energy and momentum.
 
  • #43
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Only with a scattered photon in the picture can you conserve both energy and momentum.
Thank you vela. I shall take your word for it.
 
  • #44
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2.4 Interactions of X-Rays and Gamma Rays in Matter

X-rays and gamma rays are both high-energy photons. In the energy range
1–100 keV, these photons are usually called X-rays and above 100 keV they are
usually called gamma rays. Some authors use the term ‘gamma rays’ to refer to any
photon of nuclear origin, regardless of its energy. In these notes, I often use the term
‘gamma ray’ for any photon of energy larger than 1 keV. In the next few sections,
the interactions of gamma rays with matter are discussed.

Photoelectric effect. If a charged particle penetrates in matter, it will interact with
all electrons and nuclei on its trajectory. The energy and momentum exchanged in
most of these interactions are very small, but together, these give rise to the different processes discussed in the previous chapter. When a photon penetrates in
matter, nothing happens until the photon undergoes one interaction on one single
atom. Gamma rays can interact with matter in many different ways, but the only
three interaction mechanisms that are important for nuclear measurements are the
photoelectric effect, the Compton effect and the electron–positron pair creation.
In the photoelectric absorption process, a photon undergoes an interaction with
an atom and the photon completely disappears. The energy of the photon is used
to increase the energy of one of the electrons in the atom. This electron can either
be raised to a higher level within the atom or can become a free photoelectron. If
the energy of gamma rays is sufficiently large, the electron most likely to intervene in the photoelectric effect is the most tightly bound or K-shell electron. The
photoelectron then appears with an energy given by In this equation, Z represents the charge of the nucleus andEthe energy of the Xray. The coefficient ‘n’ varies between 4 and 5 over the energy range of interest. The photoelectric cross section is a steeply decreasing function of energy (see Fig. 2.17).
Every time the photon energy crosses the threshold corresponding to the binding
energy of a deeper layer of electrons, the cross section suddenly increases. Such
jumps in the cross section are clearly visible in Figs. 2.17 and 2.18.

Compton scattering. Compton scattering is the elastic collision between a photon
and an electron. This process is illustrated in Fig. 2.14. This is a process that can
only be understood from the point of view of quantum mechanics.
A photon is a particle with energy ω. From Eq. (1.1), we know that the photon
has an impulse momentumω/c. Energy and momentum conservation constrain the
energy and the direction of the final state photon. Using energy and momentum
conservation, it is straightforward to show that the following relation holds (see
Exercise 2):
By using straightforward energy conservation, we ignore the fact that the electrons
are not free particles but are bound in the atoms, and this will cause deviations from
the simple expression above.
The value of the Compton scattering cross section for photon collisions on free
electrons can only be derived from a true relativistic and quantum mechanical calculation. It is known as the Nishina–Klein formula (see Ref. [4] and references
therein).
Equation (2.11) gives the differential cross section for the Compton scattering
into a solid angle dΩ. Integration over all angles gives the total cross section σ.
The result of the integration is given in Ref. [4]. For energies either much larger or
much smaller than the electron mass, a simple and compact expression for the total
cross section is obtained.
In these formulas, r
0 represents the classical electron radius introduced in Sect.
1.2. We see that for photon energies below the mass of the electron, the Compton
cross section is independent of energy, and for photon energies above the electron
mass, the cross section decreases as (energy)−1
.
The Nishina–Klein formula only applies to scattering of gamma rays from free
electrons. If the photon energy is much larger than the binding energy of electrons
in atoms, the effects due to this binding are small.
If the gamma energy is small, there is a large probability that the recoil electron
remains bound in the atom after the collision. The atom as a whole takes up the
energy and the momentum transferred to the electron. In this case the interaction is
called coherent Compton scattering or Rayleigh scattering. If the Compton interaction ejects the electron from the atom, the interaction is called incoherent Compton
scattering.
The angular distribution of Compton scattering described by Eq. (2.11) is illustrated in Fig. 2.15. For photon energies much below the electron mass, the scattering
is rather isotropic and back-scattering is about as likely as scattering in the forward
direction. If the photon energy is much larger than the electron mass, the scattering
is peaked into the forward direction.

Experimental Techniques in Nuclear and Particle Physics

https://alpha.physics.uoi.gr/kokkas/books/Detectors/S.Tavernier-Experimental%20Techniques%20in%20%20Nuclear%20and%20Particle%20Physics(2010)%20.pdf [Broken]
 
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  • #45
ZapperZ
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Experimental Techniques in Nuclear and Particle Physics

https://alpha.physics.uoi.gr/kokkas/books/Detectors/S.Tavernier-Experimental%20Techniques%20in%20%20Nuclear%20and%20Particle%20Physics(2010)%20.pdf [Broken]
There's a lot of things written here that are specific to interaction of gamma photons with solids, and NOT relevant to the typical photoelectric effect that one often sees with visible light. For example, the interaction of the photon in the visible light on solids is NOT with "an atom". The interaction is with the conduction electron, which does not belong to any particular atom.

Not sure why this is being posted now to a thread that's a year old.

Zz.
 
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