Ch.7 - Quantum MechanicsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: The second line of the Balmer series occurs at wavelength of 486.13 nm. To which transition can we attribute this line?a) n = 6 to n = 2b) n = 5 to n = 2c) n = 4 to n = 2d) n = 3 to n = 2e) it is to t

Problem

The second line of the Balmer series occurs at wavelength of 486.13 nm. To which transition can we attribute this line?

a) n = 6 to n = 2

b) n = 5 to n = 2

c) n = 4 to n = 2

d) n = 3 to n = 2

e) it is to the n = 1 level