An iron wire has a cross-sectional area of 7.90 10 -6 m 2 . Carry out steps (a)

Question

An iron wire has a cross-sectional area of

7.90 10 -6 m 2 . 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/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) If the wire carries a current of 9.0 A, calculate the drift speed of conduction electrons. 4.21*10**-18 Your response differs significantly from the correct answer. ??? m/s

I understand most of it except I have no idea how to solve "e". Anyone who shows the work and correct amswer will get an A+

Explanation / Answer

a ) 1 mole of iron contains 55.85g = 0.05585 kg Thus atomic weightof iron = 0.05585 kg /mol b) Density of iron is ? = 7784 kg /m3 Then molar density n/V = Density of iron / Atomic weight = 7784 kg /m3 / 0.05585 kg /mol = 1.39373*105mol/m3 c) Now number density of iron atoms is N/V =nNa/V = (1.39373*105mol/m3)(6.023 *1023atoms/mol) =8.39 *1028atoms /m3 d) Now number density of conduction electrons is NC/V = 2 * N/V = 2 * 8.39*1028electrons /m3 =1.678*1029electrons/m3 e) Now cross-sectional area A = 1.03 10-5 m2 Current I = 24.0A Then currentdensity J = I/A = 24.0A / 1.0310-5 m2 =23.3 *105A/m2 We know thatcurrent density J =(NC/V)evd 23.3*105 A/m2= (1.678*1029electrons/m3)* (1.6 *10-19C)*vd Solve for drift speed vd .

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