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