Exam 2 - Solutions

# Exam 2 - Solutions - PHYSICS 6 Exam 2(2006 SOLUTIONS I 1b...

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PHYSICS 6 Exam 2 (2006) SOLUTIONS I. 1b ; 2d; 3d; 4b; 5d; 6d II. a) λ 1 = 2L =2 × 10 -7 m λ 2 = 10 -7 m b) p 1 = h / λ 1 =(4.1 × 10 -15 eV sec)/(2 × 10 -7 m)=2.05 × 10 -8 eV sec/m p 2 = h / λ 2 = 2 × p 1 =4.1 × 10 -8 eV sec/m c) E 1 =p 1 2 /2m =( p 1 c) 2 /(2mc 2 ) =(2.05 × 10 -8 eV sec/m × 3.0 × 10 8 m/s) 2 ÷ (2 × 5.1 × 10 5 eV) = 3.71 × 10 -5 eV E 2 =4 × E 1 =14.8 × 10 -5 eV d) E(photon)= E 2 – E 1 = 1.1 × 10 -4 eV = hf λ = c/f = hc/E(photon) = 4.1x10 -15 eV s x 3x10 8 m/s ÷ 1.1 × 10 -4 eV = 1.1 x10 -2 m III. a) probability for one number 3 in one roll P = 1/6 = 0.167 or 16.7% b) prob of getting 4 in one roll = 1/6 So prob of 3 or 4 = 1/6 + 1/6 = 1/3 c) prob for no number 3 in one roll = (1-P) = 5/6 = 0.833 or 83.3% d) prob for EXACTLY TWO number 3’s in 4 rolls: There are 6 combinations that have two number 3’s and two not: (yes-yes-no-no), (yes-no-yes-no), (yes-no-no-yes),
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• Spring '06
• GaryR.Goldstein
• Physics, 2m, 2 m, 2L

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