# s01x3aans - 1 sin 8 m a a = 5.7 ×10-7 m = 0.57 m c m 10...

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PHYS-1200 PHYSICS II QUIZ 3 APRIL 11, 2001 SOLUTIONS PART A. 1. E 7. A 2. D 8. C 3. A 4. A 5. D 6. A PART B . 1. a) The answer can be read off the graph, where the stopping potential equals zero. f c = 6.0 ×10 14 Hz b) When the stopping potential equals zero, 0 = hf c Φ , so Φ = hf c . Φ = (6.63 ×10 -34 J·s)(6.0 ×10 14 Hz) Φ = 4.0 ×10 -19 J = 2.5 eV 2. a) From the symmetry of the graph, it is clear that 1/3 of the area under the plotted curve is in the range from x = 0 to x = L /3. Each of the three peaks is equal. 3 / 0 2 3 | | L dx ψ = 1/3 = 0.33 b) I deduced the answer based on the symmetry of the wave function.

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PART C. 1. a) d sin θ = m λ , so ° × = = - sin1.0 m) 10 0 . 5 ( 1 sin 8 m d d = 2.9 ×10 -6 m = 2.9 μ m b) a sin = m , so ° × = = - sin5.0 m) 10 0 . 5
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Unformatted text preview: ( 1 sin 8 m a a = 5.7 ×10-7 m = 0.57 m c) m 10 5.0 m/s) 10 . 3 s)( J 10 63 . 6 ( 8 8 34--× × ⋅ × = = = hc hf E ph E ph = 4.0 ×10-18 J = 25 eV d) 2 8 31 2 34 2 2 2 m) 10 kg)(5.0 10 2(9.11 s) J 10 63 . 6 ( 2 2---× × ⋅ × = = = m h m p E el E el = 9.65 ×10-23 J = 6.0 ×10-4 eV = 0.60 meV e) ∆ y = D tan = (0.50 m) tan 1.0° ∆ y = 8.7 ×10-3 m = 8.7 mm 2. a) For an open pipe, L v n f o 2 = . Then, ) Hz 690 ( 2 m/s 343 1 2 = = f v n L L = 0.25 m = 25 cm b) For a closed pipe, ) Hz 690 ( 2 1 2 1 2 2 1 4 = = = = o f L v n L v n f f = 345 Hz...
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s01x3aans - 1 sin 8 m a a = 5.7 ×10-7 m = 0.57 m c m 10...

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