Exam 1-solutions - Version 902 Exam 1 Mccord(50970 This...

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Version 902 – Exam 1 – Mccord – (50970) 1 This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH301 c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10 - 34 J · s m e = 9 . 11 × 10 - 31 kg R = 2 . 178 × 10 - 18 J N A = 6 . 022 × 10 23 mol - 1 1 atm = 1.01325 × 10 5 Pa 1lbs = 453 . 6g 1 gal = 3.785 L 1 in = 2.54 cm c = λν E = λ = h p = h mv 1 2 mv 2 = - Φ E n = -R parenleftbigg 1 n 2 parenrightbigg - ¯ h 2 2 m d 2 ψ d x 2 + V ( x ) ψ = ψ n ( x ) = parenleftbigg 2 L parenrightbigg 1 2 sin parenleftBig nπx L parenrightBig E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , · · · 001 10.0 points If a particle is confined to a one-dimensional box of length 300 pm, for Ψ 3 the particle is most likely to be found at 1. 300 pm. 2. 0 pm. 3. 17.3 pm. 4. 50, 150, and 250 pm, respectively. cor- rect 5. 100 and 200 pm, respectively. Explanation: 002 10.0 points Set(s) of possible values of m are A) - 4; - 3; - 2; - 1; 0; +1; +2; +3; +4 B) - 3; - 2; - 1; 0; +1; +2; +3 C) - 2; - 1; 0; +1; +2 D) - 1; 0; +1 E) 0 Select the best choice for n = 3. 1. only C 2. A, B, C, D, and E 3. only D 4. only D and E 5. only A 6. only E 7. only B, C, D, and E 8. only B 9. only C, D, and E correct Explanation: For any value of n , the possible values of range from 0 up to ( n - 1). In turn, for
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Version 902 – Exam 1 – Mccord – (50970) 2 any value of , the possible values of m range from = - , ..., 0, ..., + . Since n = 3, can be 0, 1, or 2. Therefore when = 2, C) m = - 2; - 1; 0; +1; +2 when = 1, D) m = - 1; 0; +1 when = 0, E) m = 0 003 10.0 points An electron in a 3 d orbital could have which of the following quantum numbers? 1. n = 2; = 2; m = 2 2. n = 3; = 0; m = 0 3. n = 3; = 1; m = - 1 4. n = 2; = 3; m = 0 5. n = 3; = 3; m = 1 6. n = 3; = 2; m = 0 correct 7. n = 3; = 2; m = - 3 Explanation: 3 refers to the principal quantum number n . d corresponds to the subsidiary quantum number = 2. Since = 2, m could be - 2 , - 1 , 0 , 1 or 2. 004 10.0 points In a many-electron atom, any two electrons 1. have the same energy. 2. have a different energy. 3. can have either the same or a different energy. correct Explanation: Energy is determined by distance. Elec- trons can have the same energy (distance from the nucleus) but have different sets of quan- tum numbers. 005 10.0 points Calculate the longest-wavelength line in the Balmer series for hydrogen. 1. 1875 nm 2. 486 nm 3. 657 nm correct 4. 182 nm 5. 122 nm Explanation: ν = R parenleftbigg 1 n 2 1 - 1 n 2 2 parenrightbigg = (3 . 29 × 10 15 Hz) parenleftbigg 1 2 2 - 1 3 2 parenrightbigg = 4 . 56944 × 10 14 Hz λ = c ν = 3 . 0 × 10 8 m / s 4 . 56944 × 10 14 Hz = 6 . 56535 × 10 - 7 m = 656 . 535 nm 006 10.0 points Write an equation for the second ionization energy of calcium. 1. Ca + (g) Ca 2+ (g) + e - correct 2. Ca + (g) + e - Ca(g) 3. Ca + (g) + 2 e - Ca 2+ (g) 4. Ca(g) Ca 2+ (g) + 2 e - 5. Ca + (g) + e - Ca - (g) Explanation: Definition 007 0.0 points EXTRA CREDIT
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  • Fall '07
  • Fakhreddine/Lyon
  • Atomic orbital, He+, C6 H5 Cl

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