At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 × 107 (ω∙m)–1 and 0.0012 m 2/v∙s, respectively. (a) compute the number of free electrons per cubic meter for aluminum at room temperature. (b) what is the number of free electrons per aluminum atom? assume a density of 2.7 g/cm3.

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

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