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Unformatted text preview: Version 733 – 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. 001 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 D 2. only A 3. only C 4. A, B, C, D, and E 5. only D and E 6. only B 7. only B, C, D, and E 8. only C, D, and E correct 9. only E Explanation: For any value of n , the possible values of ℓ range from 0 up to ( n 1). In turn, for 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 = 0 002 10.0 points Consider the radial distribution function (RDF) plot shown below. Which of the orbitals given as choices below would corre spond to the RDF shown. r 4 πr 2 R 2 1. 5 d correct 2. 5 p 3. 4 d 4. 4 f 5. 4 s 6. 6 d Explanation: The plot has 2 radial nodes (spherical nodes). The number of spherical nodes will al ways be equal to n ℓ 1. Only the 5 d orbital fits this case. Other orbitals that WOULD fit this would have been the 3 s , 4 p , and 6 f . 003 10.0 points What is the electronic configuration of Hg? 1. [Xe] 6 s 3 4 f 14 5 d 9 2. [Xe] 6 s 2 4 f 14 5 d 10 correct 3. [Rn] 6 s 2 4 f 14 5 d 10 4. [Kr] 5 s 3 4 d 9 5. [Xe] 6 s 2 5 d 10 Explanation: 004 10.0 points An electron is confined to a onedimensional box of length L. It falls from the second en ergy level to the ground state, and releases a photon with a wavelength of 322 nm. What is Version 733 – Exam 1 – Mccord – (50970) 2 the length of the box? 1. 292 nm. 2. 1.61 nm. 3. 0.151 nm. 4. 1.14 nm. 5. 322 nm. 6. 4.32 nm. 7. 0.303 nm. 8. 0.541 nm. correct 9. 2.92 nm. Explanation: 005 10.0 points How many electrons could be described by the quantum numbers n = 5 and m s = + 1 2 ? 1. 25 correct 2. 2 3. 1 4. 50 5. 5 Explanation: For a given principle energy level, n , a total of 2 n 2 electrons are described. Exactly half of those, n 2 , will have a + 1 2 spin. 006 10.0 points The higher the energy of electromagnetic ra diation, the 1. shorter its wavelength. correct 2. higher the velocity of light. 3. longer its wavelength. 4. lower its frequency. 5. greater its mass. Explanation: For electromagnetic radiation, the energy of radiation is related to its frequency by the equation E = h ν , where h is Planck’s constant, ν = c λ , and c is the speed of light in a vacuum (also a constant). Thus, the equation that relates energy to wavelength for electromagnetic radiation is E = h c λ As energy increases, ν (frequency) must also increase, and λ (wavelength) must decrease....
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This note was uploaded on 10/04/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.
 Fall '07
 Fakhreddine/Lyon

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