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# HW2answers - Cheung Anthony Homework 2 Due midnight Inst...

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Cheung, Anthony – Homework 2 – Due: Sep 19 2006, midnight – Inst: McCord 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Remember the first 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 m v 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 + E kinetic = 7 . 00161 × 10 - 19 J + 3 . 93936 × 10 - 17 J = 4 . 00937 × 10 - 17 J Since c = ν λ , E = h ν = h c λ λ = h c E = 6 . 626 × 10 - 34 m 2 · kg / s 4 . 00937 × 10 - 17 J × 3 . 0 × 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 m v = 6 . 626 × 10 - 34 J · s (275 kg) 125000 m 3600 s = 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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Cheung, Anthony – Homework 2 – Due: Sep 19 2006, midnight – Inst: McCord 2 c = ν λ , so ν = R 1 n 2 2 - 1 n 2 1 c λ = R 1 n 2 2 - 1 n 2 1 c = R 1 n 2 2 - 1 n 2 1 λ λ = c R 1 n 2 2 - 1 n 2 1 = 3 × 10 8 m / s (3 . 29 × 10 15 1 / s) 1 4 - 1 16 = 4 . 86322 × 10 - 7 m = 486 . 322 nm 004 (part 2 of 3) 10 points What is the name given to the spectroscope series to which this transition belongs? 1. Pfund series 2. Balmer series correct 3. Paschen series 4. Brackett series 5. Lyman series Explanation: 005 (part 3 of 3) 10 points In what region will the light lie?
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HW2answers - Cheung Anthony Homework 2 Due midnight Inst...

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