chapt 3 - PHYS 133 WEEK 3 CHAPTER 28 28.3. Solve: Using...

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PHYS 133 WEEK 3 CHAPTER 28 28.3. Solve: Using Equation 28.3 and Table 28.1, the electron current is i = nAv d = 5.9 ! 10 28 m " 3 ( ) # 0.5 ! 10 " 3 m ( ) 2 5.0 ! 10 " 5 m / s ( ) = 2.3 ! 10 18 s " 1 The time for 1 mole of electrons to pass through a cross section of the wire is t = N A ! 1 mole i = 6.02 ! 10 23 2.3 ! 10 18 s " 1 = 2.62 ! 10 5 s # 3.03 days Assess: The drift speed is small, and Avogadro’s number is large. A time of the order of 3 days is reasonable. 28.7. Solve: (a) Each gold atom has one conduction electron. Using Avogadro’s number and n as the number of moles, the number of atoms is N = nN A = m M A N A = ! V M A N A = !" r 2 L ( ) M A N A The density of gold is ρ = 19,300 kg/m 3 , the atomic mass is M A = 197 g mol 1 , r = 0.5 × 10 3 m, L = 0.1 m, and N A = 6.02 × 10 23 mol 1 . Substituting these values, we get N = 4.63 × 10 21 electrons. (b) If all the electrons are transferred a charge of (4.63 × 10 21 )( 1.60 × 10 19 C) = 740.8 C will be delivered. To deliver a charge of 32 nC, however, the electrons within a length l have to be delivered. Thus, l = ! 32 " 10 ! 9 C ! 740.8 C 10 cm ( ) = 4.32 " 10 ! 10 cm = 4.32 " 10 ! 12 m 28.11. Model: A battery is a charge escalator. Solve:
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chapt 3 - PHYS 133 WEEK 3 CHAPTER 28 28.3. Solve: Using...

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