Asked by Jeff
A 0.62-mm-diameter copper wire carries a tiny current of 2.0 μA . The molar mass of copper is 63.5 g/mole and its density is 8900 kg/m3. NA=6.02×1023
Estimate the electron drift velocity. Assume one free electron per atom.
Estimate the electron drift velocity. Assume one free electron per atom.
Answers
Answered by
Damon
2*10^-6 Coulombs/second pass a point on the wire every second
1 electron is 1.6*10^-19 Coulombs
so
2*10^-6 Coulombs/second * 1 electron/1.6*10^-19 Coulombs
= 1.25*10^13 electrons/second
electrons/s = electrons/cm^3 * area*velocity =
what is electrons/cm^3?
density = 8900*10^3 g/10^6cm^3 = 8.9 g/cm^3
1 electron/atom
so 8.9 g/cm^3 * 6*10^23 electrons/mol * 1 mol/63.5 g
= .841 * 10^23 electrons/cm^3
so
1.25*10^13 electrons/s = .841*10^23 electrons/cm^3 * pi d^2/4 * V
v is in cm/s of course, divide by 100 for m/s
1 electron is 1.6*10^-19 Coulombs
so
2*10^-6 Coulombs/second * 1 electron/1.6*10^-19 Coulombs
= 1.25*10^13 electrons/second
electrons/s = electrons/cm^3 * area*velocity =
what is electrons/cm^3?
density = 8900*10^3 g/10^6cm^3 = 8.9 g/cm^3
1 electron/atom
so 8.9 g/cm^3 * 6*10^23 electrons/mol * 1 mol/63.5 g
= .841 * 10^23 electrons/cm^3
so
1.25*10^13 electrons/s = .841*10^23 electrons/cm^3 * pi d^2/4 * V
v is in cm/s of course, divide by 100 for m/s
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