 Solution

Choose longitudinal current along $x$, magnetic field along $z$, and define the oriented Hall coefficient by $R_H=E_y/(j_x{B}_z)$. For one carrier type with signed charge $q$, a stationary transverse force balance is $q(\mathbf{E}+\mathbf{v}\times \mathbf{B}{)}_y=0$. Since $(\mathbf{v}\times \mathbf{B}{)}_y=-v_x{B}_z$, this gives $E_y=v_x{B}_z$. Using $j_x=nq{v}_x$,This is the weak-field isotropic one-carrier approximation with Hall scattering factor one, not a general multiband identity.

Under the stipulated electron-dominated approximation, the given Hall magnitudes implyTake the ratio of the intrinsic activation law:Solving gives the model's activation-gap estimateThis result uses the supplied Hall-temperature ratio and the fixed-gap, fixed-prefactor assumptions. A temperature-dependent gap or changing Hall factors would modify its interpretation.

The source's description of electrons as intrinsic silicon's majority carriers needs qualification. Intrinsic neutrality gives equal electron and hole concentrations, not an electron density majority. Electrons can nevertheless dominate conductivity and the Hall response because their mobility is larger. For two weak-field carriers,At $n=p=n_i$ this becomes $R_H=-({\mu }_e-{\mu }_h)/[e{n}_i({\mu }_e+{\mu }_h)]$. Thus ${\mu }_e>{\mu }_h$ explains the negative sign, while the one-electron formula is quantitatively accurate only if the hole contribution can be neglected. Equality $n=p$ itself is a concrete counterexample to a literal electron-number-majority claim. The numerical deductions above retain the approximation explicitly requested by the source.