exam 1 b - Version 232 Exam 1 McCord(53130 This print-out...

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Version 232 – Exam 1 – McCord – (53130) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH301tt This exam is only for McCord’s Tues/Thur CH301 class. c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10 - 34 J · s m e = 9 . 11 × 10 - 31 kg N A = 6 . 022 × 10 23 mol - 1 ν = R 1 n 2 y - 1 n 2 x where R = 3 . 29 × 10 15 s - 1 ψ n ( x ) = 2 L 1 2 sin n π x L E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , · · · 001 10.0 points A quantum mechanical particle in a box in its ground state is most likely to be found 1. in the middle of the box. correct 2. is equally likely to be found at all positions except the very center of the box. 3. at both edges of the box. 4. is equally likely to be found at all positions in the box. 5. at the very edge of the box. Explanation: The ground state is Ψ 1 which has maximum probability in the middle of the box. 002 10.0 points The graph shows the radial distribution plots for the 1s wavefunctions for H, He, and He + . A B C radius 4 π r 2 Ψ 2 Which plot is the 1s wavefunction for the He + ion? 1. There is no way to know 2. C 3. A correct 4. B Explanation: H has one electron and one proton. He has two electrons and two protons. He + has one electron and two protons. Therefore the elec- tron will have the greatest attraction to the the nuclei with two protons. In He there will be a slight reduction in the e ff ective nuclear charge due to the electron-electron repulsion. In He + , there is only one electron so the radius will be the smallest. Plot A is peaked closest to the nucleus so it will be the smallest. 003 10.0 points What are the correct quantum numbers ( n , , m ) for the seventh electron of Cl? 1. 2, 1, 1 correct 2. 2, 2, 1 3. 3, 2, 1 4. 3, 1, 0 5. 2, 1, 2
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Version 232 – Exam 1 – McCord – (53130) 2 6. 3, 1, 1 7. 2, 0, 1 8. 3, 0, 0 Explanation: The electron configuration for Cl is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 5 The seventh electron is in a 2 p orbital, so n = 2, = 1, m = - 1, 0, or +1. 2, 1, 0 and 2, 1, - 1 are not given as choices. 004 10.0 points Carbon emits photons at 745 nm when ex- posed to blackbody radiation. How much energy would be obtained if 44 g of carbon were irradiated? 1. 2 . 67 × 10 - 19 J 2. 7 . 08 × 10 6 J 3. 9 . 11 × 10 - 21 J 4. 1 . 17 × 10 - 17 J 5. 7 . 08 × 10 3 J 6. 5 . 90 × 10 5 J correct Explanation: λ = 745 nm = 7 . 45 × 10 - 7 m m C = 44 g Assume each carbon atom emits one pho- ton. For each photon E 1 = h ν = h c λ = (6 . 626 × 10 - 34 J · s) (3 × 10 8 m / s) 7 . 45 × 10 - 7 m = 2 . 66819 × 10 - 19 J , so the total energy emitted is E = E 1 n = 2 . 66819 × 10 - 19 J C atom × 6 . 022 × 10 23 C atoms 1 mol C atoms × 1 mol C atoms 12 . 01 g C × (44 g C) = 5 . 88663 × 10 5 J . 005 10.0 points How many electrons can possess this set of quantum numbers: principal quantum number n = 4, magnetic quantum number m = - 1? 1. 14 2. 12 3. 16 4. 8 5. 0 6. 4 7. 2 8. 18 9. 10 10. 6 correct Explanation: Use the rules for the quantum numbers: If n = 4 then = 0 , 1 , 2 , 3; however, for m = - 1, = 1 , 2 , 3. Each of these permit- ted sets of values of n , and m specifies ONE orbital: n = 4, = 1, m = - 1: 4 p n = 4, = 2, m = - 1: 4 d n = 4, = 3, m = - 1: 4 f and each orbital can have m s = ± 1 2 ; i.e. , can hold two electrons.
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