Question

An iron wire has a cross-sectional area of 5.70 ✕ 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 20.0 A, calculate the drift speed of conduction electrons. ______m/s

Answer 1

a)The atomic mass of iron is
M = 55.8 g
1 mole of iron contains 55.8 g = 0.0558 kg.
So there is 0.0558 kg of iron in 1 mole. 0.0558 kg/mol
b) The density of iron is
= 7,874 kg /m³
We will first find the molar volume, which is the volume occupied by one mole of material.
molar volume is the ratio of molar mass to the mass density.
m_v = m /
m_v = 0.0558 kg/m³ / 7874 kg
m_v =  7.086 x 10^-6 m³
The molar density = 1 / 7.086 x 10^-6 =  141111.1 mol/m³
c) The number density of iron
1 mole = 6.023 x 10²³ atoms
N = molar density x no. of atoms in 1 mole
N = 141111.1 mol/m³ x 6.023 x 10²³ =  8.499 x 10²⁸ atoms/m³
d) The drift velocity v_d = I / n q A
n is the number of free electrons per m³, q is the charge of electron.
n = 2 x N = 2 x  8.499 x 10²⁸ atoms/m³ = 16.998 x 10²⁸ electrons/m³
v_d = 20.0 A / (16.998 x 10²⁸ x 1.6 x 10^-19 C x 5.70 x 10^-6 m²)
v_d =  1.29 x 10^-4 m/s