Answer:
a) 1.98 *10^29 electrons /m³
b) 3.285 electrons / aluminium atom
Explanation:
Step 1: Data given
At room temperature the electrical conductivity = 3.8 × 10^7 (Ω*m)^–1
the electron mobility for aluminum = 0.0012 m² / V*s
a) Calculate the number of free electrons per cubic meter for aluminum at room temperature
n = σ/eµ
with σ = the electrical conductivity of aluminium = 3.8 × 10^7 (Ω*m)^–1
with e = the elementary charge of an electron = 1.6 *10^-19
with µ = the electron mobility for aluminum = 0.0012 m² / V*s
n = 3.8*10^7 / ((1.6*10^-19)(0.0012))
n =1.98 *10^29 electrons /m³
(b) What is the number of free electrons per aluminum atom?
Number of atoms of aluminium per cubic meter
N = (Na*ρ)/MM
with Na = the number of avogadro = 6.022 * 10^23
with ρ = the density = 2.7 g/cm³
with MM = Molar mass of aluminium = 26.98 g/mol
N = ((6.022 * 10^23)*(2.7*10^6 ))/26.98 g/mol
N = 6.027 * 10^28 atoms / m³
Number of free electrons per aluminium atom = 1.98 *10^29 / 6.027 * 10^28
Number of free electrons per aluminium atom = 3.285 electrons / aluminium atom
