CH 301 Exam 1- Atomic Theory 9-16-10

CH 301 Exam 1- Atomic Theory 9-16-10 - Version 733 – Exam...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 733 – 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. 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 one-dimensional 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....
View Full Document

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.

Page1 / 9

CH 301 Exam 1- Atomic Theory 9-16-10 - Version 733 – Exam...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online