 Solution

In the simple transport model, ${\mu }_e=e{\tau }_e/m_e^{\ast }$ and ${\mu }_h=e{\tau }_h/m_h^{\ast }$. Conduction bands in many common semiconductors have lighter transport masses than the dominant valence hole bands. For comparable momentum-relaxation times, this gives larger electron mobility. It is a common trend, not a theorem: band anisotropy, light and heavy hole populations and different scattering times can alter the comparison. The hole's mass is not the mass of an atom, and crystal translational symmetry does not make its mobility infinite.

For the intrinsic germanium sample, electrons and holes both contribute positive conventional conductivity:ThereforeThis is the density of each species; the combined carrier number density is $2n_i$. Assigning separate densities from the two mobilities would violate intrinsic charge neutrality.