4
$\begingroup$

I am trying to understand diagrams that involve showing phonon scattering processes that contribute to raman spectroscopy peaks.

For example, I drew the one at the bottom of this post. This is supposed to the scattering process responsible for the D peak in graphene. I'm trying to understand exactly what the arrows represent.

At initial glance it looks almost as if the phonon scatters and increases energy of magnitude $E_L$ and transitions to another branch, then scatters again, losing energy of magnitude $D$, which is somehow indicative of the D peak?

enter image description here

Thanks!

$\endgroup$

2 Answers 2

3
$\begingroup$

The vertical axis is energy and the horizontal axis is k-vector (crystal momentum).

$E_L$ is the energy of the incoming photon, not the phonon.

The vertical arrow on the left describes the process where light excites an electron. Photons have almost no momentum (on the scale of this plot), so the arrow is essentially vertical.

The downward-tilting right arrow describes the process where the electron scatters by emitting a phonon, with energy D. The phonon has a lot of momentum, so by conservation of momentum the electron momentum has to change by a large (equal and opposite) amount.

The rest of the process is not shown. I bet that the electron elastically scatters (off a defect) back to the left side of the diagram (a horizontal left arrow), then emits a photon (a vertical downward arrow), returning to its original state.

The Raman shift is the energy D (or in terms of wavenumbers, it's D divided by Planck's constant). This is showing the Stokes-shift Raman process. The anti-Stokes process would involve an upward-tilting right arrow instead of a downward-tilting one.

$\endgroup$
1
$\begingroup$

Just to add some more notes to Steve's response:

  • This process only involves one $D$ phonon, which has energy approx 1335 cm${}^{-1}$ and is located at a Kohn anomaly at the $K$-point of the Brillouin zone.
  • Note that for this particular Raman process with the $D$ peak, you have to go outside of the first Brillouin zone to an adjacent one.
  • If the graphene is pristine you will not have a $D$ peak, instead you will have two $D$ phonons which give rise to the 2D peak (which has energy 2670 cm${}^{-1}$). The reason there are two involved is to conserve momentum in the process. If you have a $D$ peak in your Raman spectrum it is an indication of defects in your sample.
  • It's called a $D$ phonon because there is a similar process in diamond, but in that case the $D$ phonon is at the $\Gamma$ point and not the $K$ point.
  • Note that it's actually very relevant/important to note that the laser first excites an electron to the conduction band, and THEN the phonon scattering occurs. In a "conventional Raman scattering process" you have no electronic transition, but only vibrational or rotational. But the fact that there is an electronic transition enables you to probe electronic properties of graphene with Raman, which is VERY significant to be able to do!
  • To get a better idea of Raman peaks in graphene, a good starting point is Ferrari and Basko 2013 which has a nice diagram of many of the graphene Raman scattering processes, and Malard et al. 2009 which also has a great plot of the graphene phonon disperion relation so you can see the origin of the peaks.
  • There is also a book covering a lot of stuff on this called Raman Spectroscopy in Graphene Related Systems.

Good luck.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.