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17 July, 14:12

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.

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  1. 17 July, 14:13
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    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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