 Solution

Put $x={\mu }_{\mathrm{B}}B/(k_{\mathrm{B}}T)$. Summing the two Boltzmann factors gives the occupied-site internal partition functionIf the single atom can occupy any of the $M$ identical sites, its complete surface partition function isOne also commonly calls the one-site factor $z$ itself $q_{\mathrm{ads}}$. Both conventions are valid, but the combinatorial factor in part (c) must use the same convention. We use the complete one-atom surface function $q_{\mathrm{ads}}=Mz$ below and give the equivalent one-site expression.