# hw4 - 4-1 F corresponds to the charge passed to deposit one...

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4-1 F corresponds to the charge passed to deposit one mole of monovalent element at a cathode. As one mole contains Avogadro’s number of atoms, e = 96 500 C 6.02 10 23 = 1.60 10 # 19 C . 4-2 (a) Total charge passed = i * t = 1.00 A ( ) 3 600 s ( ) = 3 600 C . This is 3 600 C 1.60 " 10 # 19 C = 2.25 " 10 22 electrons. As the valence of the copper ion is two, two electrons are required to deposit each ion as a neutral atom on the cathode. The number of Cu atoms = number of electrons 2 = 1.125 " 10 22 Cu atoms. (b) So the weight (mass) of a Cu atom is: 1.185 g 1.125 " 10 22 atoms = 1.05 " 10 # 22 g . (c) m = q molar weight 96 500 2 ( ) or molar weight = m 96 500 ( ) 2 q = 1.185 g ( ) 96 500 C ( ) 2 3 600 C = 63.53 g . 4-3 Thomson’s device will work for positive and negative particles, so we may apply q m " V # B 2 ld . (a) q m " V # B 2 ld = 2 000 V ( ) 0.20 radians 4.57 \$ 10 % 2 T ( ) 2 0.10 m ( ) 0.02 m ( ) = 9.58 \$ 10 7 C kg (b) As the particle is attracted by the negative plate, it carries a positive charge and is a proton. q m p = 1.60 " 10 # 19 C 1.67 " 10 # 27 kg = 9.58 " 10 7 C kg \$ % & ( ) (c) v x = E B = V d B = 2 000 V 0.02 m 4.57 " 10 # 2 T ( ) = 2.19 " 10 6 m s (d) As v x ~ 0.01 c there is no need for relativistic mechanics. 4-6 The velocity of fall v = " y " t = 0.004 m 15.9 s = 2.52 # 10 \$ 4 m s . (a) The radius, a , is given by a = 9 " v 2 # g \$ % & ( ) 1 2 = 9 1.81 * 10 + 5 kg m s ( ) 2.52 * 10 + 4 m s ( ) 2 800 kg m 3 ( ) 9.81 m s 2 ( ) \$ % & & ( ) ) 1 2 = 1.62 * 10 + 6 m .

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The mass, m , is given by m = " V = " 4 3 # \$ % & ( ) a 3 = 1.33 ( ) 800 kg m 3 ( ) ) ( ) 1.62 * 10 + 6 m ( ) 3 = 1.42 * 10 + 14 kg .
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