# Compton Scattering in Substances Other Than Graphite

It shows that as a function of scattering angle the scattered photons lose energy and therefore have a longer wavelength (lambda prime).

There are some number of scattered photons that still have their original incident wavelength (lambda). This is because they collided with an electron that was still strongly bound to its parent atom and so the photon didn't pass any of its energy as it would to a free electron.

These results are for Compton's experiment where the photon target was Graphite. Do we know if this experiment was tested on any target other than Graphite in which the electrons were more or less strongly bound to the parent atom and see how the results differ?

There are some number of scattered photons that still have their original incident wavelength (lambda). This is because they collided with an electron that was still strongly bound to its parent atom and so the photon didn't pass any of its energy as it would to a free electron.

I slightly disagree with the explanation given in the video (linked below in your comment).

The left peak in all the sub-figures of your image (with the unchanged wavelength $$\lambda$$ and with angle $$\theta$$) are not photons scattered by an electron, but photons scattered by a nucleus.

The shift of wavelength for scattering by an electron is given by Compton's formula: $$\lambda'-\lambda=\frac{h}{m_ec}(1-\cos\theta) =2.4\cdot 10^{-12}\text{ m}\cdot(1-\cos\theta)$$ The atomic binding energy of the electron (even for a stronger bound inner electron) is neglectable. See the last section of hte answer.

And the shift of wavelength for scattering by a nucleus is much smaller due to the bigger mass $$m_C$$ of the carbon nucleus: $$\lambda'-\lambda=\frac{h}{m_Cc}(1-\cos\theta) =0.00011\cdot 10^{-12}\text{ m}\cdot(1-\cos\theta)$$

These results are for Compton's experiment where the photon target was Graphite. Do we know if this experiment was tested on any target other than Graphite in which the electrons were more or less strongly bound to the parent atom and see how the results differ?

Compton's experiments is done with high-energy photons (X-rays or gamma rays), i.e. with photon energies higher than $$1$$ keV. This is much higher than the atomic binding energy of the electron. E.g. the binding energy of an electron in graphite is only a few eV (even for the electrons in the more tightly bound inner shells). Therefore the atomic binding energy of the electrons can be safely neglected when compared to the energy the X-ray photon.

As far as I know, Compton's experiment has been repeated also with metals and non-metals. And it gives essentially the same results as with graphite. This was to be expected by the reasoning above.

• Thank you Thomas. Yes I had got this information from a video: youtube.com/watch?v=rGy7nsC8O_Y (at about 11:50). If what you say is correct and the first peak is just from photons passing through the graphite without interaction, then why are those photons being detected at 90 degrees, 135 degrees etc? Would they not ONLY be at 0 degrees. – user1551817 Jan 24 at 18:49
• @user1551817 Thanks for the video-link. It seems I misunderstood the meaning of the figures in your question. I have corrected my answer. – Thomas Fritsch Jan 24 at 20:08