Question

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 .

Answer 1

a) 1.98 *10^29 m³

b) 3.28 free electrons per aluminum atom

Step-by-step explanation:

Step 1: Data given

The electrical conductivity for aluminium = 3.8*10^7 (Ω*m)^–1

The electron mobility for aluminum = 0.0012 m²/V*s

Density = 2.7 g/cm³

Step 2: The number of free electrons can be given as:

n = σ/ e*µ

with σ = The electrical conductivity for aluminium = 3.8*10^7 (Ω*m)^–1

with e = 1.602 *10^-19

with µ = The electron mobility for aluminum = 0.0012 m²/V*s

n = (3.8*10^7)/(1.602 *10^-19 *0.0012)

n =1.98 *10^29 m³

Step 3: Calculate number of free electrons per aluminium atom

N = Na*ρ / M

with Na = The constant of avogadro = 6.022 *10^23 atoms /mol

ρ = Density = 2.71 g /cm³*10^6 cm³ /m³

with M = molar mass of Aluminium = 26.98 g/mol

N = (6.022 *10^23 *2.71 g /cm³*10^6 cm³ /m³)/ 26.98 g/mol

N = 6.03*10^28 m^-3

To find the number of free electrons per aluminum atom we calculate:

n/N = 1.98*10^29 m^−3 /6.03*10^28 m^-3 = 3.28