1141_Physics ProblemsTechnical Physics

1141_Physics ProblemsTechnical Physics - 482 P40.53...

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Unformatted text preview: 482 P40.53 Introduction to Quantum Physics (a) At the top of the ladder, the woman holds a pellet inside a small region ∆xi . Thus, the uncertainty principle requires her to release it with typical horizontal momentum 2H 1 , so ∆p x = m∆v x = . It falls to the floor in a travel time given by H = 0 + gt 2 as t = 2 g 2 ∆x i the total width of the impact points is bg ∆x f = ∆xi + ∆v x t = ∆xi + A= where 2m A 2H = ∆x i + g ∆x i i 2H . g d i =0 d b ∆x g d ∆x f To minimize ∆x f , we require FG IJ 2 m∆ x K H 1− or i A =0 ∆xi2 ∆x i = A . so The minimum width of the impact points is d∆x i = FGH ∆x + ∆A IJK x f min i i =2 A = ∆xi = A L 2e1.054 6 × 10 J ⋅ sj OP L 2a2.00 mf O ∆x i =M d MN 5.00 × 10 kg PQ MNM 9.80 m s PQP 12 −34 (b) FI GH JK 2 2H mg f −4 min 2 14 . 14 = 5.19 × 10 −16 m Additional Problems P40.54 ∆VS = FG h IJ f − φ H eK e From two points on the graph FG h IJ e4.1 × 10 Hzj − φ H eK e φ F hI 3.3 V = G J e12 × 10 Hzj − . H eK e 14 0= 14 and Combining these two expressions we find: FIG. P40.54 (a) φ = 1.7 eV (b) h = 4. 2 × 10 −15 V ⋅ s e (c) At the cutoff wavelength e je hc λc =φ = λ c = 4.2 × 10 −15 V ⋅ s 1.6 × 10 −19 C FG h IJ ec H eK λ c e3 × 10 m s j a1.7 eVfe1.6 × 10 j J eVj = 8 −19 730 nm ...
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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