Question

An iron wire has a cross-sectional area of 5.80 ✕ 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 12.0 A, calculate the drift speed of conduction electrons. m/s

Answer 1

We know
atomic mass of Fe = 55.8
which means 1 mole contains 55.8 grams or =0.0558 kg
=0.0558 kg/mole >> (a)

"Starting with the density of iron and the result of part (a) the number of moles of iron per cubic meter.
density iron 7870 kg/m³
0.0558 kg / 7870 kg/m³ = 7.09e-6m³ (b) which is 1 mole
1 mole / 7.09e-6m³ = 141000 mole/m³ >>(c)

(d)
"number density of iron atoms"

1 mole = 6.022×10^23 molecules
141000 mole/m³ x 6.022×10^23 atoms/mole = 8.49e28 atoms/m³

"two conduction electrons per iron atom"
=2*8.49*10^28
=1.70e29 electons/m³

(e)
drift velocity Vd
Vd = I / nqA
Vd is drift velocity in m/s
I is current
n is the number of charge carriers per m³
for copper, 8.5e28 electrons per m³.
A = the cross sectional area =5.8*10^-6
q is the charge of the charge carriers (electrons)
q = –1.602e-19 Coulomb (charge on an electron)

Vd = (12) / (1.70e29)(1.602e-19)(5.8*10^-6)
=0.00007596m/s