= Solution
At $B_z=8\,\mathrm T$,
$$
\boxed{\nu_8=\frac{N_Sh}{8e}=\frac54\nu_{10}.}
$$
The cyclotron spacing is $\hbar e(8\,\mathrm T)/m^*$, 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.
\b[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 $\nu_8\simeq2.5$, while $N_S=3.87\times10^{15}\,\mathrm{m^{-2}}$ instead gives $\nu_8\simeq2$. 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.
Back to article page
