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## 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

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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

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Chemistry
3 years
2021-08-22T17:00:56+00:00
2021-08-22T17:00:56+00:00 1 Answers
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## Answers ( )

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