HW2answers pg1 - E kinetic = 7 00161 × 10-19 J 3 93936 ×...

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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Remember the frst exam is on Wednesday, 9/20. CHECK the website to see which room to go to. Some oF you will take the exam in HMA (Hogg Memorial Auditorium) and some will take the exam in our classroom. CHECK THE CLASS WEB SITE For this inFormation. You MUST go to the right room. 001 (part 1 oF 1) 10 points The work Function For chrominum metal is 4.37 eV. What wavelength oF radiation must be used to eject electrons with a velocity oF 9300 km / s? Correct answer: 4 . 95788 nm. Explanation: v = 9300 km / s = 9 . 3 × 10 6 m / s The wavelength oF radiation needed will be the sum oF the energy oF the work Function plus the kinetic energy oF the ejected elctron. E work function = (4 . 37 eV) × (1 . 6022 × 10 - 19 J / eV) = 7 . 00161 × 10 - 19 J E kinetic = 1 2 mv 2 = 1 2 (9 . 10939 × 10 - 31 kg) × (9 . 3 × 10 6 m / s) 2 = 3 . 93936 × 10 - 17 J E total = E work function
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Unformatted text preview: + E kinetic = 7 . 00161 × 10-19 J + 3 . 93936 × 10-17 J = 4 . 00937 × 10-17 J Since c = ν λ , E = hν = hc λ λ = hc E = 6 . 626 × 10-34 m 2 · kg / s 4 . 00937 × 10-17 J × 3 . × 10 8 m / s = 4 . 95788 × 10-9 m × 10 9 nm 1 m = 4 . 95788 nm 002 (part 1 oF 1) 10 points Calculate the wavelength oF a motorcycle oF mass 275 kg traveling at a speed oF 125 km/hr. 1. 1.93 × 10-38 m 2. 2.41 × 10-36 m 3. 1.93 × 10-41 m 4. 6.94 × 10-38 m correct 5. 2.08 × 10-29 m Explanation: v = 125 km/h = 125000 m/h m = 275 kg λ = h mv = 6 . 626 × 10-34 J · s (275 kg) p 125000 m 3600 s P = 6 . 93923 × 10-38 m 003 (part 1 oF 3) 10 points Use the Rydberg Formula For atomic hydrogen to calculate the wavelength For the transition From n = 4 to n = 2. 1. 205 nm 2. 8.63 nm 3. 486 nm correct 4. 2.45 nm 5. 94 . 9 nm Explanation:...
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