Exam 1 - Version 089 – Exam 1 – mccord –(51600 1 This...

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Unformatted text preview: Version 089 – Exam 1 – mccord – (51600) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10 − 34 J · s m e = 9 . 11 × 10 − 31 kg R = 3 . 29 × 10 15 s − 1 R = 2 . 178 × 10 − 18 J N A = 6 . 022 × 10 23 mol − 1 g = 9 . 81 m/s 2 E = hν c = λ/ν λ = h p = h mv 1 2 mv 2 = hν- Φ E n =- R n 2 (hydrogen atom) ν = R parenleftbigg 1 n 2 1- 1 n 2 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 An electron is confined in a 1-dimensional box 3 nm long. Calculate the energy required for it to move from the n = 8 to the n = 9 energy level. 1. 3 . 414 × 10 − 28 J 2. 1 . 717 × 10 − 14 J 3. 2 . 133 × 10 − 14 J 4. 5 . 238 × 10 − 19 J 5. 2 . 133 × 10 − 19 J 6. 1 . 138 × 10 − 19 J correct 7. 6 . 694 × 10 − 21 J Explanation: n 1 = 8 n 2 = 9 h = 6 . 626 × 10 − 34 J · s L = 13 × 10 − 9 m m = 9 . 109 × 10 − 31 kg Δ E = E n 2- E n 1 = ( n 2 2- n 2 1 ) h 2 8 mL 2 = (9 2- 8 2 ) (6 . 626 × 10 − 34 J · s) 2 8 (9 . 109 × 10 − 31 kg) (3 × 10 − 9 m) 2 = 1 . 13802 × 10 − 19 J 002 10.0 points Bohr’s theory of the hydrogen atom assumed that 1. electromagnetic radiation is given off when the electrons move in an orbit around the nucleus. 2. because a hydrogen atom has only one electron, the emission spectrum of hydrogen should consist of only one line. 3. the electron can exist in any one of a set of discrete states (energy levels) and can move from one to another by emitting or absorbing radiation. correct 4. the electron in a hydrogen atom can jump from one energy level to another without gain or loss of energy. 5. energy, in the form of radiation, must Version 089 – Exam 1 – mccord – (51600) 2 be continually supplied to keep the electron moving in its orbit. Explanation: 003 10.0 points How many d electrons does Cr have? 1. 3 2. 5 correct 3. 4 4. 6 5. 2 Explanation: The electron configuration for Cr is [Ar] 3 d 5 4 s 1 which shows that it has 5 elec- trons in the d-orbital. 004 10.0 points An electron in a 3 d orbital could have which of the following quantum numbers? 1. n = 2; ℓ = 3; m ℓ = 0 2. n = 3; ℓ = 3; m ℓ = 1 3. n = 3; ℓ = 2; m ℓ =- 3 4. n = 3; ℓ = 1; m ℓ =- 1 5. n = 3; ℓ = 0; m ℓ = 0 6. n = 3; ℓ = 2; m ℓ = 0 correct 7. n = 2; ℓ = 2; m ℓ = 2 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....
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Exam 1 - Version 089 – Exam 1 – mccord –(51600 1 This...

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