Version PREVIEW – Quiz 1 – JOHNSON – (53755) 1This print-out should have 8 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.LDE Q01 03001 10.0 pointsWhich of the following types of electromag-netic radiation has the shortest wavelength?1. 3.57 × 10−19J correct2. 2.74 × 10−19J3. 3.12 × 10−19J4. 3.05 × 10−19J5. 2.83 × 10−19JExplanation:LDE Q01 02002 10.0 pointsWhich of the following types of electromag-netic radiation has the shortest wavelength?1. Infrared2. Ultraviolet3. Microwave4. Gamma correct5. RadioExplanation:LDE Q01 06003 10.0 pointsShow that the wave nature of particles is neg-ligible for macroscopic objects by calculat-ing the de Brogli e wavelength of a 2 g pap erairplane thrown at Dr. Laude traveling at5 m · s−1.1. 1.656 × 10−33m2. 6.626 × 10−32m correct3. 6.037 × 1032m4. 1.509 × 1034m5. 2.650 × 10−34mExplanation:2 g = 2 × 10−3kgλ =hmv=6.626 × 10−34J · s(2 × 10−3kg) (5 m · s−1)= 6.626 × 10−32mLDE uncertainty calculation 001004 10.0 pointsA proton in the form of a hydrogen ion hasa fairly well defined position with an uncer-tainty of only 10−11m. What would theminimum uncertainty in the proton’s velocitybe? (A proton has a mass of 1.672×10−27kg)1. 4, 950 m · s−12. 9, 900 m · s−13. 19, 800 m · s−14. 3, 150 m · s−1correct5. 31, 100 m · s−1Explanation:∆x(m∆v) ≥h4π∆v ≥h4πm∆x≥6.626 × 10−34J · s4π (1.672 × 10−27kg) (10−11m)≥ 3, 153 m · s−1LDE quantum rules005 10.0 pointsWhich of the following sets of quantum num-bers are valid, i.e. don’t vi olate any boundaryconditions?Version PREVIEW – Quiz 1 – JOHNSON – (53755) 2I) n = 3, ℓ = 2, mℓ= −2, ms= +12II) n = 9, ℓ = 5, mℓ= 6, ms= +12III) n = 2, ℓ = 1, mℓ= 0, ms= +1IV) n = 2, ℓ = 0, mℓ= 0, ms= +121. II, III2. I, II, IV3. IV only4. I, III, IV5. II only6. III only7. I, IV correct8. I onlyExplanation:Set II and III are invalid. For II, mell= 6is disallowed because ℓ = 5. For III, ms= +1is disallowed because msmay only be +12or−12.LDE Aufbau, Hund, Pauli 002006 10.0 points3p ↑↑↑ ↑3s ↑↓2p ↑↓ ↑↓ ↑↓2s ↑↓1s ↑↓Consider the electron filling diagram fora ground state atom above. Assume anyunwritten orbitals are empty. Which of thefollowing does it v iolate?I) The Aufbau principleII) Hund’s ruleIII) The Pauli exclusion principle1. III only correct2. II only3. II, III4. I, II, III5. I, II6. I only7. I, IIIExplanation:The Pauli exclusion principle dictates thata given electron and a given set of quantumnumbers have a 1 to 1 rela tion. The twoelectrons in first 3p orbital have the same spinand thus the same set o f quantum numbers, inviolation of the the Pauli exclusion principle.Msci 05 1412007 10.0 pointsHow many electrons can be in the n = 2 shell?1. 22. 323. 184. 8 correctExplanation:LDE classical failure 001008 10.0 pointsWhich of the following statement(s) is/aretrue about the photoelectric effect?I) Classical physics failed to explain someof the observed phenomena.II) Given light of high enough int ensity, elec-trons can be ejected off any surface.III) Einstein employed the concept that pho-tons have q uantized amo unts of energy toexplain the effect.1. I, III correct2. II, IIIVersion PREVIEW – Quiz 1 – JOHNSON – (53755) 33. I only4. I, II, III5. I, IIExplanation:Classical mechanics predicted that light ofany wavelength would be able to eject elec-trons from a metal surface if it was sufficientlyintense, which was inconsistent with the ob-served threshold effect. This threshold effect,the ejection energy, required that light energywas
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