PC Chapter 04 - 4 The Particle Nature of Matter 4-1 F...

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47 4 The Particle Nature of Matter 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, - = = × × 19 23 96 500 C 1.60 10 C 6.02 10 e . 4-2 (a) Total charge passed ( 29 ( 29 = = = * 1.00 A 3 600 s 3 600 C i t . This is - = × × 22 19 3 600 C 2.25 10 1.60 10 C 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 = = × 22 number of electrons 1.125 10 Cu 2 atoms. (b) So the weight (mass) of a Cu atom is: - = × × 22 22 1.185 g 1.05 10 g 1.125 10 atoms . (c) ( 29 = molar weight 2 96 500 m q or ( 29 ( 29 ( 29 = = = 2 2 molar weight 96 500 1.185 g 96 500 C 63.53 g 3 600 C m q . 4-3 Thomson’s device will work for positive and negative particles, so we may apply θ 2 q V m B ld . (a) ( 29 ( 29 ( 29( 29 θ - = = × × 7 2 2 2 0.20 radians 2 000 V 0.10 m 0.02 m 9.58 10 C kg 4.57 10 T q V m B ld (b) As the particle is attracted by the negative plate, it carries a positive charge and is a proton. - - × = = × × 19 7 27 1.60 10 C 9.58 10 C kg 1.67 10 kg p q m
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48 CHAPTER 4 THE PARTICLE NATURE OF MATTER (c) ( 29 - = = = × = × 2 6 2 000 V 4.57 10 T 2.19 10 m s d 0.02 m x E V v B B (d) As ~ 0.01 x v c there is no need for relativistic mechanics. 4-4 ( 29 ( 29 ( 29 ( 29 ( 29    = = = =       2 2 2 2 1 1 1 1 1 2 2 2 2 y e x e x e x F l Ee l V e l y a t m v m v d m v ( 29 ( 29 θ   = = = = =      = + = +  2 2 1 2 2 1 tan 2 y y x x e x x e x e x Dv Da t Ee l V e l y D D D v v m v v d m v V e l l y y y d m D v Solving for e e m yields ( 29 = + 2 2 x v I l e d e y V D m or ( 29 = + 2 1 2 x e yv d e m Vl D . 4-5 (a) Draw the curved path of the electron as follows: ( 29 - + = 2 2 2 r y l r - + + = + = 2 2 2 2 2 2 2 2 r ry y l r l y r y y r l r r – y (b) ( 29 - = = × = × = 4 8 1 060 V 2.5 10 m 2.39 10 m s 0.795 0.017 7 V v c Bd . Using the classical expression = = p mv Bre gives ( 29 ( 29 - - × = = × × = 8 31 19 2.39 10 m s 9.11 10 kg 1.60 10 C 0.076 9 m 0.017 7 T mv r Be . To find the value of y corresponding to this r use - + = 2 2 2 0 y ry l or - - + × = 2 4 0.153 8 6.10 10 0 y y . Using the quadratic formula, ( 29 ( 29 ( 29 ( 29 - - = - ± - × = + ± = 1 2 2 1 2 2 4 4 2 0.153 8 4 6.101 10 0.153 8 0.150 m 2 b ac y b a y - × = 3 4.08 10 m 0.004 08 m (the first answer is discarded because r y ). Using the relativistic expression γ = = p mv Bre gives
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MODERN PHYSICS 49 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 γ - - × × = = = - × 31 8 1 2 19 2 1 9.11 10 kg 2.39 10 m s 0.126 7 m 1 0.795 0.017 7 T 1.60 10 C me r Be . The y values corresponding to this r value are found from - - + × = 2 4 0.253 6.10 10 0 y y or = 0.251 m y , 0.002 43 m . Discarding the first answer because r must be greater than y , leaves = 0.002 43 y m in good agreement with ( 29 = ± observed 0.002 4 0.000 5 m y . 4-6 The velocity of fall - = = = × 4 0.004 m 2.52 10 m s 15.9 s y v t .
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