Past exam of the quantum condensed matter course of the University of Cambridge / 2022 / Section A / 3

An angle-resolved photoemission spectroscopy intensity map locates occupied electronic band structure through its bright dispersing ridges. For a nearly isotropic graphene cone,Fit straight lines to the reasonably linear portions of the ridges, avoiding the broad central crossing, and extrapolate them to locate the Dirac point energy $E_D$. The PDF places the crossing near $k_y=0$ and plotted energy roughly $1.4$–$1.5\mathrm{eV}$. On a lower branch, a representative change from approximately $(|k_y|,E)=(0.10{\mathring{\mathrm{A}}}^{-1},0.9\mathrm{eV})$ towards $(0,1.5\mathrm{eV})$ gives a slope near $6\mathrm{eV}\mathring{\mathrm{A}}$. Allowing for the ridge width gives about $5$–$7\mathrm{eV}\mathring{\mathrm{A}}$, henceHere $1\mathrm{eV}\mathring{\mathrm{A}}/\hslash =1.52\times 10^5\mathrm{m}{\mathrm{s}}^{-1}$.

To determine the carrier Fermi energy, identify the Fermi level from a calibrated photoemission cutoff, then measure its separation from the Dirac point:The separation of the two Fermi crossings gives $2k_F$ for a cut through the cone centre. This determines the doping energy relative to the Dirac point; the absolute plotted energy origin is not itself a Fermi energy.

The reproduced panel has no marked Fermi level or stated energy-zero calibration. Its upper ridges end near $1.9\mathrm{eV}$, but that alone does not prove that the endpoint is a Fermi cutoff. If it is the intended cutoff, the estimate is $E_F^{\mathrm{cone}}\simeq 0.4$–$0.5\mathrm{eV}$ and $k_F\sim 0.06$–$0.08{\mathring{\mathrm{A}}}^{-1}$. If instead the plotted numbers are binding energies measured from an unstated Fermi zero, the inferred separation would be roughly $1.4$–$1.5\mathrm{eV}$. The supplied figure securely supports the slope estimate, but it does not distinguish these energy-reference conventions; an unconditional numerical Fermi energy requires the missing calibration.