Answer
$\frac{n}{N_{Al}} = 3.29$
Work Step by Step
Given: electrical conductivity of Aluminum= 3.8 x $10^{7}$ $(Ωm)^{-1}$ electron mobility of Aluminum = 0.0012 $m^{2}$/ Vs
Required: number of free electrons per aluminum atom assuming density of 2.7 $\frac{g}{cm^{3}}$
Solution: Determine the number of aluminum atoms per cubic meter, $N_{Al}$ using Equation 4.2 and using the atomic weight of Aluminum of 26.98 g/mol:
$N_{Al} = \frac{N_{A}p^{'}}{A_{Al}} = \frac{(6.022 x 10^{23})(2.7 \frac{g}{cm^{3}})(10^{6} \frac{cm^{3}}{m^{3}})}{(26.98 g/mol)} = 6.026 \times 10^{28} m^{-3}$
Computing for the number of free electrons per aluminum atom:
$\frac{n}{N_{Al}} = \frac{1.98 \times 10^{29} m^{-3}}{6.026 \times 10^{28} m^{-3}} = 3.29$
