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Unformatted text preview: Version 902 – Exam 1 – Mccord – (50970) 1 This printout should have 31 questions. Multiplechoice 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 = hν λ = h p = h mv 1 2 mv 2 = hν Φ E n =R parenleftbigg 1 n 2 parenrightbigg ¯ h 2 2 m d 2 ψ d x 2 + V ( x ) ψ = Eψ 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 onedimensional 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 ℓ 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 ℓ range from 0 up to ( n 1). In turn, for Version 902 – Exam 1 – Mccord – (50970) 2 any value of ℓ , the possible values of m ℓ range from = ℓ , ..., 0, ..., + ℓ . Since n = 3, ℓ can be 0, 1, or 2. Therefore when ℓ = 2, C) m ℓ = 2; 1; 0; +1; +2 when ℓ = 1, D) m ℓ = 1; 0; +1 when ℓ = 0, E) m ℓ = 0 003 10.0 points An electron in a 3 d orbital could have which of the following quantum numbers? 1. n = 2; ℓ = 2; m ℓ = 2 2. n = 3; ℓ = 0; m ℓ = 0 3. n = 3; ℓ = 1; m ℓ = 1 4. n = 2; ℓ = 3; m ℓ = 0 5. n = 3; ℓ = 3; m ℓ = 1 6. n = 3; ℓ = 2; m ℓ = 0 correct 7. n = 3; ℓ = 2; m ℓ = 3 Explanation: 3 refers to the principal quantum number n . d corresponds to the subsidiary quantum number ℓ = 2. Since ℓ = 2, m ℓ could be 2 , 1 , , 1 or 2. 004 10.0 points In a manyelectron 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 longestwavelength 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 . × 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....
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This note was uploaded on 12/06/2010 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

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