e1f07 - CePrNdPmSmEuGdTbDyHoErTmYbLu...

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Unformatted text preview: CePrNdPmSmEuGdTbDyHoErTmYbLu ThPaUNpPuAmCmBkCfEsFmMdNoLr HHe LiBeBCNOFNe NaMgAlSiPSClAr KCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKr RbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXe CsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRn FrRaAcRfDbSgBhHsMt 12 345678910 1112131415161718 192021222324252627282930313233343536 373839404142434445464748495051525354 555657727374757677787980818283848586 878889104105106107108109 5859606162636465666768697071 90919293949596979899100101102103 1.00794.0026 6.9419.012210.81112.01114.006715.999418.998420.1797 22.989824.305026.981528.085530.973832.06635.452739.948 39.098340.07844.955947.8850.941551.996154.938055.84758.933258.6963.54665.3969.72372.6174.921678.9679.90483.80 85.467887.6288.905991.22492.906495.94(98)101.07102.9055106.42107.8682112.411114.82118.710121.75127.60126.9045131.39 132.9054137.327138.9055178.49180.9479183.85186.207190.2192.22195.08196.9665200.59204.3833207.2208.9804(209)(210)(222) (223)(226)(227)(261)(262)(263)(262)(265)(266) 140.115140.9076144.24(145)150.36151.965157.25158.9253162.50164.9303167.26168.9342173.04174.967 232.0381231.0359238.0289(237)(244)(243)(247)(247)(251)(252)(257)(258)(259)(260) 1A8A 2A3A4A5A6A7A 3B4B5B6B7B8B1B2B 118 21314151617 3456789101112 Periodic Table of the Elements Version 001 Exam 1 David Laude (54705) 2 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. V1:1, V2:1, V3:1, V4:1, V5:2. ACAMP 02 0001 08:02, general, multiple choice, > 1 min, fixed. 001 (part 1 of 1) 6 points A 200 nm photon has how many times the energy of a 700 nm photon? 1. 3.5 correct 2. 4.2 3. 0.29 4. 0.24 5. 9 . 93 10- 19 6. 2 . 84 10- 19 Explanation: Energy of Light: E = hc For the 200 nm photon: E = hc = (6 . 626 10- 34 J s)(3 10 8 m s- 1 ) 200 10- 9 m = 9 . 94 10- 19 J For the 700 nm photon: E = hc = (6 . 626 10- 34 J s)(3 10 8 m s- 1 ) 700 10- 9 m = 2 . 84 10- 19 J Thus 9 . 94 10- 19 J 2 . 84 10- 19 J = 3 . 5 ChemPrin3e T01 26 08:05, general, multiple choice, < 1 min, fixed. 002 (part 1 of 1) 6 points Which of the following emission lines corre- sponds to part of the Balmer series of lines in the spectrum of a hydrogen atom? A) n = 2 n = 1 B) n = 4 n = 2 C) n = 4 n = 1 D) n = 3 n = 2 E) n = 4 n = 3 1. B and D only correct 2. A, D, and E only 3. A and C only 4. E only 5. B and C only 6. D and E only 7. B, C, and E only Explanation: The Balmer series is produced by elec- tronic transitions which either begin (absorp- tion spectra) or end (emission spectra) at the energy level n = 2. These correspond mostly to the visible region. ChemPrin3e 01 30 08:05, general, numeric, > 1 min, normal....
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