= Solution
Under the one-carrier <Drude model>, the magnitude of the <Hall coefficient> gives an effective <carrier density>
$$
n_H=\frac1{e|R_H|}=2.60\times10^{28}\,\mathrm{m^{-3}}.
$$
Convert the material <mass density> to $1850\,\mathrm{kg\,m^{-3}}$ and its <molar mass> to $0.009\,\mathrm{kg\,mol^{-1}}$. The atomic <number density> is
$$
n_{\rm at}=\frac{1850}{0.009}N_A=1.238\times10^{29}\,\mathrm{m^{-3}},\qquad
\boxed{\frac{n_H}{n_{\rm at}}=0.210\ \text{effective carriers per atom}}.
$$
Here $N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}$ 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.
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