This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Cheung, Anthony Homework 2 Due: Sep 19 2006, midnight Inst: McCord 1 This printout should have 30 questions. Multiplechoice 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 = , E = h = h c = h c 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 = 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: Cheung, Anthony Homework 2 Due: Sep 19 2006, midnight Inst: McCord 2 c = , so = R 1 n 2 2 1 n 2 1 c = R 1 n 2 2 1 n 2 1 c = R 1 n 2 2 1 n 2 1 = 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?...
View
Full
Document
This note was uploaded on 11/11/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Fakhreddine/Lyon
 Chemistry

Click to edit the document details