f01x3aan - W 5.0- = = = r P A P I I = 6.4 10 2 W/m = 640...

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NAME PHYS-1200 PHYSICS II QUIZ 3 NOVEMBER 14, 2001 SOLUTIONS PART A. 1. D 6. D 2. C 7. E 3. D 8. A 4. D 5. C PART B . 1. a) m 10 00 . 5 m 10 00 . 8 7 6 1 - - × × = = λ L N N 1 = 16 b) ) 16 )( 25 . 1 ( / 2 = = = L n n L N N 2 = 20 c) THEY ARE IN PHASE . N 2 N 1 = 6, which is an integer. 2. a) The answer can be read off the graph, where the stopping potential equals zero. f c = 3.0 ×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)(3.0 ×10 14 Hz) Φ = 2.0 ×10 -19 J = 1.2 eV
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NAME PART C . 1. a) Since θ is small, the position of the first minimum is given by, d D r λ 22 . 1 = , so m 10 5.0 m) m)(10 10 (6.0 22 . 1 22 . 1 2 7 - - × × = = r D d d = 1.5 ×10 -4 m = 0.15 mm b) 2 2 2 m) 10 5.0 (
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Unformatted text preview: W 5.0- = = = r P A P I I = 6.4 10 2 W/m = 640 W/m c) P = Nhf , so m/s) 10 3.0 ( s) J 10 6.63 ( m) 10 W)(6.0 (5.0 8 34 7 = = =--hc P hf P N N = 1.5 10 19 photons/s d) INCREASES THE SIZE OF THE CENTRAL MAXIMUM DECREASES, AND THE POWER REMAINS THE SAME 2. a) For an open pipe, L v n f o 2 = . Then, ) Hz 570 ( 2 m/s 343 1 2 = = f v n L L = 0.30 m = 30 cm b) For a closed pipe, ) Hz 570 ( 2 1 2 1 2 2 1 4 = = = = o f L v n L v n f f = 285 Hz...
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This test prep was uploaded on 04/18/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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f01x3aan - W 5.0- = = = r P A P I I = 6.4 10 2 W/m = 640...

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