An iron wire has a cross-sectional area equal to 1.00 10-5 m2. Carry out the fol

Question

An iron wire has a cross-sectional area equal to 1.00 10-5 m2. Carry out the following steps to determine the drift speed of the conduction electrons in the wire if it carries a current of 21.0 A.
(a) How many kilograms are there in 1.00 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/m^3
(c) Calculate the number density of iron atoms using Avogadro's number. atoms/m^3
(d) Obtain the number density of conduction electrons given that there are two conduction electrons per iron atom. electrons/m^3
(e) Calculate the drift speed of conduction electrons in this wire. m/s

Explanation / Answer

a.) We need to use molar mass (which is usually represented as gram/mol) and convert it into kg/mol. You can find molar mass of iron on any periodic table on the internet

so 55.8 g/mol x 1kg/1000g = 0.0558kg/mol

b.) So we have to grab the density of iron and make sure it has a value the relates to kg and use our kg molar mass to find it.

S o based on wikipedia we find that iron's density is 7.874 g/cm3 = 0.007874 kg/cm3

So we can multiply the density and molar mass (we can flip one of them upside down so that the units cancel out to give the desired outcome, if you have any questions please comment)

0.007874 kg/cm3 x (1mol/0.0558kg) = 0.1411 mol/cm3

since the question asks for m3 we have to convert it

0.1411 mol/cm3 x (100cm/1m)3 = 141100 mol/m3

c.) Avogadro's number is 6.02 x 1023 atom/mol

141100mol/m3 x 6.02x1023 atom/mol = 8.49 x 1028 atom/m3

d.) Since there are 2 conducting electrons per atom we just have to multiply it by 2 and we'll have

1.70 x 1029 electrons/m3

e.) Now that we have the number of electrons for number of volume of wire we can figure out the drift speed of the conduction electrons in the wire.

Since the cross section of the wire is 1x10-5m2 a meter of the wire would have a volume of

1x10-5m2 x 1 m = 1x10-5m3

From here we can calculate the number of electrons that are in here.

1.70 x 1029 x 1x10-5m3 = 1.70 x 1024 electrons

If we multiply the number of electrons by their elementary charge we can figure out the Amp and the Coulombs and the speed at which the electrons in the wire is drifting.

1.70 x 1024 electrons X 1.60x10-19C/electron = 2.72 x 105 C

21 Amp = 2.72x105C/ (x seconds)

X seconds = 1.30x104 seconds

1m/1.30x104 seconds = 7.72x10-5m/sec

That would be the drift speed of the electrons

Hope this helps :)

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