hw4_solutions

hw4_solutions - PHY 251 Fall 2009 homework problem set 4...

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PHY 251 Fall 2009: homework problem set 4, due in the PHY 251 drop box in room A-129 by noon on Friday, Oct. 9. 1. Do Serway problem 4.2 on p. 146. This problem illustrates how electrochemistry experi- ments already gave some idea about the discrete charge of the electron, once you believe in discrete atoms and know Avogadro’s number! Answer: Copper sulfate involves Cu +2 and SO 2 4 , so there are two valence electrons per atom. We then find that the fraction of a mole is (1 C/sec ) · (3600 sec ) (96 , 500 C/mol ) · 2 = 0 . 0187 mol . The number of coppper atoms deposited is then (0 . 0187 mol ) · (6 . 02 × 10 23 atoms/mol ) = 1 . 13 × 10 22 , the weight of a copper atom is 1 . 185 × 10 3 kg 1 . 13 × 10 22 atoms = 1 . 05 × 10 25 kg , and the molar mass of copper is 1 . 185 g 0 . 0187 mol = 63 . 4 g/mol which is about right (of course they gave us numbers that would work out!). 2. Do Serway problem 4.3. Answer: In the apparatus, we have F y = q V d = m Δ v y Δ t and we also have v x = ℓ/ Δ t or Δ t = ℓ/v x . Therefore we can write Δ v y = qV Δ t md = qV md v x . We can then figure out the angle θ : tan θ = v y v x = qV ℓ mdv 2 x so q m = dv 2 x tan θ V ℓ . Now from the magnetic field cancelling the electric field we can say qV d = qv x B so v x = V Bd = 2000 4 . 57 × 10 2 · 0 . 02 = 2 . 19 × 10 6 m/s which means relativistic effects are small ( γ = 1 . 000 027 ). We can also substitute this result for velocity into our expression for q/m : q m = d V ℓ V 2 B 2 d 2 tan θ = V ldB 2 tan θ = 2000 0 . 10 · 0 . 02 · (4 . 57 × 10 2 ) 2 tan(0 . 2) = 9 . 7 × 10 7 C/kg . A proton has q/m = 1 . 6 × 10 19 / 1 . 67 × 10 27 = 9 . 7 × 10 7 so it seems like we have a good candidate . . . 1

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3. Do Serway problem 4.4. Answer: The electric field the electron experiences when it travels between the two plates is E y = V/d , and the acceleration it experiences is a = F/m = qE/m = qV/ ( md ) . Since we expect v y v x , we can say that the time that the electron experiences this acceleration is Δ t = ℓ/v x . Consequently the electron receives a velocity kick in the ˆ y direction of v y = a y · Δ t = qV md · v x so that its angle θ leaving the field region is θ = v y /v x = y 2 /D (the latter by geometry in the small angle limit) or θ = y 2 D = v y v x = qV md v 2 x which can be solved for y 2 to give y 2 = qV Dℓ mdv 2 x .
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hw4_solutions - PHY 251 Fall 2009 homework problem set 4...

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