 Solution

Under the one-carrier Drude model, the magnitude of the Hall coefficient gives an effective carrier densityConvert the material mass density to $1850\mathrm{kg}{\mathrm{m}}^{-3}$ and its molar mass to $0.009\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. The atomic number density isHere $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$ is the Avogadro constant. The positive Hall coefficient gives a hole-like response in a one-carrier interpretation. The Hall carrier count versus valence distinction matters: beryllium has two valence electrons, but the result is neither two free Electrons nor a literal fractional atom donating charge. Its energy bands and Fermi surface invalidate the simple isotropic one-carrier inference. Electron-like and hole-like contributions with different charge carrier mobilities can yield this sign and effective count.