Compute the number of electrons that each lead atom donates, on average, to a bulk piece of lead metal. Room temperature data for lead: The

Question

Compute the number of electrons that each lead atom donates, on average, to a bulk piece of lead metal. Room temperature data for lead: The conductivity of lead is 4.90 × 104 1/(Ω·m) The electron mobility of lead is 2.3 cm2/(V·s) The mass density of lead is 11.4 g/cm3 The atomic weight of lead is 207 g/mol

in progress 0
Hải Đăng 3 years 2021-08-22T17:00:56+00:00 1 Answers 16 views 0

Answers ( )

  1. diemkieu
    0
    2021-08-22T17:02:33+00:00
    Reply

    Answer:

    4 electrons/atom

    Explanation:

    The conductivity of the lead σ = neμ where n = electron density, e = electron charge = 1.602 × 10⁻¹⁹ C and μ =  electron mobility of lead = 2.3 cm²/(V·s) = 2.3 × 10⁻⁴ m²/(V.s)

    Making n subject of the formula, we have

    n = σ/eμ

    Since σ = 4.90 × 10⁶ (Ω·m)⁻¹

    Substituting the values of the variables into the equation, we have

    n = σ/eμ

    n = 4.90 × 10⁶ (Ω·m)⁻¹/(1.602 × 10⁻¹⁹ C × 2.3 × 10⁻⁴ m²/(V.s))

    n = 4.90 × 10⁶ (Ω·m)⁻¹/(3.6846 × 10⁻²³ Cm²/(V.s))

    n = 1.33 × 10²⁹ electrons/m³

    We now find the number of moles of lead present in 1 m³ of lead.

    So n’ = ρ/M where ρ = density of lead = 11.4 g/cm³ = 11.4 g/cm³ × 10⁶ cm³/m³ = 11.4 × 10⁶ g/m³ and M = atomic weight of lead = 207 g/mol

    So, n’ = ρ/M

    N = 11.4 × 10⁶ g/m³/207 g/mol

    n’ = 0.0551 × 10⁶ mol/m³

    n’ = 5.51 × 10⁴ mol/m³

    Since n’ = N/N’ where N = number of atoms of lead in 1 m³ of lead and N = Avogadro’s constant = 6.022 × 10²³ mol⁻¹

    N = n’N’ = 5.51 × 10⁴ mol/m³ × 6.022 × 10²³ mol⁻¹

    N = 33.18 × 10²⁷ atoms/m³

    N = 3.318 × 10²⁸ atoms/m³

    So, the number of electron per atom is N” = n/N

    = 1.33 × 10²⁹ electrons/m³ ÷ 3.318 × 10²⁸ atoms/m³

    = 0.4 × 10¹ electrons/atom

    = 4 electrons/atom

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )