= Solution
For a single carrier type with unit Hall factor, the magnitude of the <Hall voltage> is $V_H=IB/(net)$, where $t$ is the slab thickness parallel to the <magnetic field>. Therefore
$$
\boxed{n=\frac{IB}{etV_H}=1.25\times10^{24}\ \mathrm{m^{-3}}.}
$$
The longitudinal <electrical resistivity> is $\rho=Rwt/L=1000(2\times10^{-3})(0.5\times10^{-3})/(5\times10^{-3})=0.20\ \Omega\mathrm m$. Using $1/\rho=ne\mu$, the <electron mobility> magnitude is
$$
\boxed{\mu=\frac1{\rho ne}=2.5\times10^{-5}\ \mathrm{m^2\,V^{-1}s^{-1}}=0.25\ \mathrm{cm^2\,V^{-1}s^{-1}}.}
$$
Without the Hall polarity these data do not identify the carrier charge sign. The inference assumes one dominant carrier population; electron-hole compensation would require a multicarrier analysis.
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