 Solution

In the one-band electron Hall effect model, $R_H=-1/(ne)$, with $e>0$. The measured Hall coefficient therefore givesA monovalent body-centered cubic lattice has two conduction electrons per conventional unit cell. Thus $n{a}^3=2$, andCounting two spin states within the Fermi sphere gives $k_F=(3{\pi }^{2}n{)}^{1/3}=9.0418\times 10^9{\mathrm{m}}^{-1}$. Using the specified effective mass in the dispersion,There is one mole of conduction electrons per mole of sodium. Its molar density of states at $E_F$ is $3N_A/(2E_F)$, givingUsing the volume density of states in c givesThe effective mass changes the dispersion and density of states; the electron Bohr magneton used for the spin coupling retains the bare electron mass. Replacing it by $e\hslash /(2m^{\ast })$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.