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## Photoelectric effect vs Compton scattering

 Quote by mike168 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.

 Quote by vela 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)?
 '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

 Quote by Dickfore For typical crystals $a \sim 5 \stackrel{o}{A}$. Take an Avogadro number of atoms, and one electron per atom ($N_e = 1$). We have: $$\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}$$
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.

 Quote by Per Oni 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...

 Quote by M Quack 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.

 Quote by Per Oni 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.

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