test 1 - Version 171 – Exam 1 – McCord –(53130 1 This...

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 171 – Exam 1 – McCord – (53130) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. McCord CH301tt This exam is only for McCord’s Tues/Thur CH301 class. c = 3 . 00 × 10 8 m/s h = 6 . 626 × 10- 34 J · s m e = 9 . 11 × 10- 31 kg N A = 6 . 022 × 10 23 mol- 1 ν = R parenleftbigg 1 n 2 y- 1 n 2 x parenrightbigg where R = 3 . 29 × 10 15 s- 1 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 Dr. Schrodinger would predict that the tran- sition from the first energy level to the second energy level would require a photon with the longest wavelength for which of the following elements? 1. K 2. Na 3. Li 4. Rb correct Explanation: Based on Schrodinger’s analysis, the energy levels become closer together as the size of the “box” increases. 002 10.0 points In the spectrum of atomic hydrogen, a violet- blue line is observed at 434 nm. What are the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line? 1. n = 3, n = 2 2. n = 6, n = 3 3. n = 4, n = 3 4. n = 5, n = 3 5. n = 5, n = 2 correct 6. n = 6, n = 2 7. n = 4, n = 2 Explanation: λ = 434 nm = 4 . 34 × 10- 7 m Because the line is in the visible part of the spectrum, it belongs to the Balmer series for which the ending n is 2. For the starting value of n , ν = c λ = 3 × 10 8 m / s 4 . 34 × 10- 7 m = 6 . 909 × 10 14 s- 1 Using the Ryberg formula, ν = (3 . 29 × 10 15 s- 1 ) parenleftbigg 1 n 2 2- 1 n 2 1 parenrightbigg ν 3 . 29 × 10 15 s- 1 = parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg 1 n 2 2 = 1 n 2 1- ν 3 . 29 × 10 15 s- 1 = 1 4- 6 . 909 × 10 14 s- 1 3 . 29 × 10 15 s- 1 = 0 . 04 n 2 2 = 25 n 2 = 5 003 10.0 points Version 171 – Exam 1 – McCord – (53130) 2 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. 50, 150, and 250 pm, respectively. cor- rect 2. 0 pm. 3. 100 and 200 pm, respectively. 4. 17.3 pm. 5. 300 pm. Explanation: 004 10.0 points What are the correct quantum numbers ( n , ℓ, m ℓ ) for the seventh electron of Cl? 1. 3, 1, 0 2. 2, 1, 2 3. 2, 2, 1 4. 3, 2, 1 5. 3, 0, 0 6. 2, 0, 1 7. 3, 1, 1 8. 2, 1, 1 correct Explanation: The electron configuration for Cl is 1 s 2 2 s 2 2 p 6 3 s 2 3 p 5 The seventh electron is in a 2 p orbital, so n = 2, ℓ = 1, m ℓ =- 1, 0, or +1. 2, 1, 0 and 2, 1,- 1 are not given as choices. 005 10.0 points How many electrons are in the third principal level ( n = 3) of a chromium (Cr) atom? 1. 13 correct 2. 4 3. 24 4. 8 5. 2 6. 12 7. 6 Explanation: There are 2 electrons in the 3 s orbital, 6 electrons in the 3 p orbitals and 5 electrons in the 3 d orbitals. Remember that Cr is an exception. One electron from the 4 s orbital gets elevated to a 3 d orbital giving it a half filled 3 d orbital....
View Full Document

This note was uploaded on 02/02/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

Page1 / 8

test 1 - Version 171 – Exam 1 – McCord –(53130 1 This...

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