# Conductivity of Wüstite

"Wüstite is often represented as $\ce{Fe_{1-x}O}$, where $x$ is some small fraction less than unity. In this nonstoichiometric $\ce{Fe_{1-x}O}$ material conduction is electronic and the material behaves as a p-type semiconductor, i.e. the $\ce{Fe^3+}$ ions act as electron acceptors, generating holes.

Determine the electrical conductivity of a specimen of wüsite that has a hole mobility of $\pu{3.1E-5 m2 V-1 s-1}$ and for which the value of $x$ is 0.030. Assume that the acceptor states are

saturated, i.e. one hole exists for every $\ce{Fe^3+}$ ion. Wüsite has a sodium chloride crystal structure with a unit cell edge length of 0.437 nm."

My Solution:

4 units of $\ce{FeO}$ for rocksalt structure but for every $\ce{Fe^2+}$ vacancy $2$ $\ce{Fe^3+}$ are needed for electroneutrality, this implies $0.003:1\rightarrow0.24:4$

$$\mathrm{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_a)(10^3))$$ $$=\pu{2.225838592E-26 kg}$$ $$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$ $$=\pu{266.7 kg m-3}$$ Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$ $$\implies\sigma=(266.7)(1.6\times10^{-19})(3.1\times10^{-5})$$ $$\approx1.3\times10^{-21}\:(\Omega.m)^{-1}$$

I have the feeling this is wrong so any help would be appreciated.

after revisiting the problem I realised where I went wrong. The main mistake with not being careful with units and not reading the qurstion and having incorrect values for hole mobility. $$\text{mass}_{\ce{Fe(III)}}=((0.24)(55.85))/((N_\mathrm{a})(10^3))$$ $$=2.225838592\times10^{-26}\: \mathrm{kg}$$ $$\rho_{\ce{Fe(III)}}=2.225838592\times10^{-26}/(4.37\times10^{-10})^3$$ $$=\mathrm{266.7\: kg\: m^{-3}}$$ $$\text{No. of holes}=\rho_{\ce{Fe(III)}}/1\:\text{AMU}_\ce{{Fe(III)}}$$ $$\Rightarrow \text{No. of holes}=266.7/(9.296\times10^{-26})$$ $$=2.868975904\times\pu{10^{27}\:holes/m^3}$$ Conductance for a p-type semiconductor is given as: $$\sigma=p(q)(\mu_{h})$$ $$\implies\sigma=(2.868975904\times10^{27})(1.6\times10^{-19})(3.1\times10^{-5})$$ $$\approx1.4\times10^{4}\:(\Omega.m)^{-1}$$