Answer
n = 1.98 x $10^{29}$ $m^{-3}$
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 $m^{3}$ for aluminum at room temperature
Solution:
Using Equation 18.8,
n = $\frac{σ}{(|e|µ_{e})}$ = $\frac{3.8 \times 10^{7} (Ωm)^{-1}}{(1.602 \times 10^{-19} C)(0.0012 m^{2}/Vs)}$
n = 1.98 x $10^{29}$ $m^{-3}$
