 Solution

The Mott two-current model treats the two spin populations as parallel conduction channels when spin-flip scattering is weak. Thus $\sigma ={\sigma }_{\uparrow }+{\sigma }_{\downarrow }$ and ${\rho }^{-1}={\rho }_{\uparrow }^{-1}+{\rho }_{\downarrow }^{-1}$. Light, broad-band carriers conduct efficiently; scattering into narrow $d$-like states relaxes their current. In a simple spin-conserving $s$–$d$ picture,In copper, the filled $d$ bands supply little final-state phase space at $E_F$, so both spin channels have low resistivity. In cobalt, the nearly filled majority $d$ band suppresses majority-channel scattering, while minority $d$ states remain available. The stated iron reversal corresponds to stronger scattering-weighted majority-channel phase space: its majority $d$ band is not closed as in cobalt. Band filling, exchange splitting, velocities and scattering matrix elements set this scattering-weighted spin-channel asymmetry; majority population alone does not select the better conductor. The given iron ordering is a material/scattering input, not a universal consequence of the two-current postulate. Different impurities and interfaces can change the asymmetry.^[ref]