Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2023 / Paper 2 / Section A(ii) / 15 / v / Solution

Using the magnitude of the classical slope,The Landau filling factor at $1\mathrm{T}$ isUnder the convention that the “last Landau level” means the last resolved spin branch, its degeneracy per area is $eB/h$. Setting ${\nu }_{\mathrm{fill}}=1$ givesAbove this threshold all carriers fit into that lowest resolved branch, which is partially filled when ${\nu }_{\mathrm{fill}}<1$.

There is a spin convention to state explicitly: if “Landau level” instead means one orbital level with unresolved twofold spin degeneracy, its capacity is $2eB/h$, and all carriers first fit into orbital index $n=0$ at $B=3.24\mathrm{T}$, where ${\nu }_{\mathrm{fill}}=2$. The slope alone determines $n_s$ and ${\nu }_{\mathrm{fill}}$, but does not specify the spin splitting.