CHAPTER27The Microscopic Theory of Electrical Conduction1* ·In the classical model of conduction, the electron loses energy on average in a collision because it loses the drift velocity it had picked up since the last collision. Where does this energy appear?The energy lost by the electrons in collision with the ions of the crystal lattice appears as Joule heat (I2R).2 ·A measure of the density of the free-electron gas in a metal is the distance rs, which is defined as the radius of the sphere whose volume equals the volume per conduction electron. (a) Show that rs= (3/4pn)1/3, where nis the free-electron number density. (b) Calculate rsfor copper in nanometers. (a) The volume occupied by one electron is 1/n= (4/3)prs3. Thus, rs= (3/4pn)1/3. (b)n= 8.47×1028m–3(See Table 27-1). Evaluate rsrs= 1.41×10–10m = 0.141 nm 3 ·(a) Given a mean free path ?= 0.4 nm and a mean speed vav= 1.17×105m/s for the current flow in copper at a temperature of 300 K, calculate the classical value for the resistivity ?of copper. (b) The classical model suggests that the mean free path is temperature independent and that vavdepends on temperature. From this model, what would ?be at 100 K?(a) Use Equ. 27-7 (b) vav∝T1/2?= 104)10(1.6108.47101.171091021928531---××××××××O.m = 0.123 μO.m ?100= (0.123 μO.m)(100/300)1/2= 0.071 μO.m 4 ·Calculate the number density of free electrons in (a) Ag (?= 10.5 g/cm3) and (b) Au (?= 19.3 g/cm3), assuming one free electron per atom, and compare your results with the values listed in Table 27-1. (a), (b) n= ?NA/M; ?= density, M= molar mass (a) nAg= 5.86×1022el./cm3; (b) nAu= 5.90×1022el./cm3. The results agree with Table 27-1 5* ·The density of aluminum is 2.7 g/cm3. How many free electrons are present per aluminum atom? ne= electrons/atom = nM/?NAne= 106.022.726.981018.12322××××= 3.00 electrons/atom 6 ·The density of tin is 7.3 g/cm3. How many free electrons are present per tin atom?
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