Question

An iron wire has a cross-sectional area of 5.90 ✕ 10⁻⁶ m². Carry out steps (a) through (e) to compute the drift speed of the conduction electrons in the wire.

(a) How many kilograms are there in 1 mole of iron?
_____________ kg/mol

(b) Starting with the density of iron and the result of part (a), compute the molar density of iron (the number of moles of iron per cubic meter).
___________mol/m³

(c) Calculate the number density of iron atoms using Avogadro's number.
__________ atoms/m³

(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom.
___________electrons/m³

(e) If the wire carries a current of 38.0 A, calculate the drift speed of conduction electrons.
___________ m/s

Answer 1

a.)

Since molar mass of iron = Mfe = 55.85 gm/mol

also , 1 gm = 1*10^-3 kg

So, Mfe = 55.85*(1*10^-3) kg/mol

Mfe = 55.85*10^-3 kg/mol

b.)

Density of iron = (rho)fe = 7.86*10^3 kg/m^3

So, molar density = (rho)fe/Mfe = (7.86*10^3 kg/m^3)/(55.85*10^-3 kg/mol)

molar density = 1.41*10^5 mol/m^3

c.)

Since number of atoms in 1 mole = Na = 6.02*10^23 atoms = Avogadro's number

So, density of atoms = Na*(molar density) = (6.02*10^23 atoms/mol)*(1.41*10^5 mol/m^3)

density of atoms = 8.49*10^28 atoms/m^3

d.)

given one iron atom have 2 conduction electrons,

So, density of charge carriers(n) = (2 electrons/atom)*(density of atoms)

n = (2)*(8.49*10^28) electrons/m^3

n = 1.70*10^29 electrons/m^3

e.)

Since current(I) = n*e*A*Vd

given I = 38.0 amp

n = 1.70*10^29

e = 1.60*10^-19 C

A = 5.90*10^-6 m^2

So, drift velocity(Vd) = I/(n*e*A)

Vd = 38.0/[(1.70*10^29)*(1.60*10^-19)*(5.90*10^-6)]

Vd = 2.37*10^-4 m/sec

Please upvote.