 Solution

At $B_z=8\mathrm{T}$,The cyclotron spacing is $\hslash e(8\mathrm{T})/m^{\ast }$, smaller than at 10 T, and each resolved level's degeneracy per area is also reduced by $4/5$. Thus the same carrier density occupies more levels. This ratio is determined without the missing graph, but it does not specify whether the chemical potential lies in a localized tail or an extended-state centre.

TODO — missing density/plateau assignment: a numerical index and chemical-potential marker at 8 T require the unavailable density information. For example the first illustrative density in part (i) gives ${\nu }_8\simeq 2.5$, while $N_S=3.87\times 10^{15}{\mathrm{m}}^{-2}$ instead gives ${\nu }_8\simeq 2$. These produce different schematic placements even though both satisfy every surviving sample parameter. Unresolved spin degeneracy and disorder add further qualifications; no unique 8 T label should be invented.