Find the number of free electrons per m3 contributing to the current. Recall tha

Question

Find the number of free electrons per m3 contributing to the current. Recall that the number of grams of an element in one mole of the element is the same as its atomic weight. One mole of copper is 63.5 g of copper. So, from the density of copper, we see that 1 cm3 contains 8.92 g of copper. So each cm3 of copper contains 8.9g / 63.5 g/mol = 0.1405 mol of copper. Recall also that 1 mol of any substance contains Avogadro's number of atoms (6.02 x 1023). So 1 cm3 of copper contains (0.1405 mol)(6.02 times 1023 atoms/mol) = 8.46 times 1022 atoms, and 1 m3 would contain 8.46 times 1028 atoms. Then the number of electrons per m3 according to the assumptions made is n - times m-3 Find the drift velocity of the electrons in the wire. Since the current of n electrons per m3 each with charge -e and with cross section A of the wire is I = -neAvd Slove for vd vd = - I /neA Substitute numerical values. vd = - I / nea = - (2.6a) / N(1.60 TIMES 10-19C)(3.6 TIMES 10-6M2) = - M/S

Explanation / Answer

You can use this example as a guide to doing the calculation yourself. Just plug in the numbers from your problem in place of the ones in the example: (Note: you will need to use 64 for the atomic weight of copper, and 9g/cm^3 for density, 9 amps for the value of L, and also re-calculate the area based on a radius of 6mm.) ________________________________ Electricity is most commonly conducted in a copper wire. Copper has a density of 8.94 g/cm³, and an atomic weight of 63.546 g/mol, so there are 140685.5 mol/m³. In 1 mole of Copper there are 6.02×10^23 atoms(Avogadro's constant). Therefore in 1m³ of copper there are about 8.5×1028 atoms (6.02×1023 x 140685.5 mol/m³. Copper has one free electron per atom, so n is equal to 8.5×1028 electrons per m³. Assume a current I=3 amperes, and a wire of 1 mm diameter (radius in meters = 0.0005m). This wire has a cross sectional area of 7.85×10-7' m2 ('A= p 0.00052 ). The charge of 1 electron is q=1.6×10-19 Coulombs. The drift velocity therefore can be calculated: V= I/nAq V = 3 / (8.5×1028 x 7.85×10-7 x 1.6×10-19 ) V = 0.00028 m/s [V] = [Amps] / [electron/m3] x [m2] x [Coulombs/electron] = [coulombs] / [seconds] x [electron/m3] x [m2] x [Coulombs/electron] = [meters] / [second] Therefore in this wire the electrons are flowing at the rate of 0.00028 m/s, or very nearly 1.0 m/hour. This is quite slow, however in 1 hour 6,75x1022 electrons will flow through the wire. If the wire were 1 meter long, (assuming 18 Gauge copper wire) and 3 amps was flowing through it, it would consume as much energy as a 150mWatt light bulb. (Very rough calculation)*

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