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: What is the smallest wavelength in the Balmer's series?

Problem

What is the smallest wavelength in the Balmer's series?

Solution

We can determine the smallest wavelength, λmin in the Balmer series using the Balmer Equation shown below:


1λmin=R×1nf2-1ni2

where: 

λmin  = wavelength, m at the highest initial principal energy level: ↑ni ↓λ

R = 1.0974 x 107m-1 (Rydberg Constant)      **value can be found in textbooks or online 
ni = initial principal energy level  
nf = final principal energy level = 2 for Balmer Series 


Recall that for the Balmer series the final principal energy level nf is always = 2. 

The smallest wavelength, λmin will be the minimum wavelength corresponding to the highest initial energy level, ni possible or approaching infinity,  (a symbol used to denote a super high number)  for a Hydrogen atom.

Recall that the highest transition releases the highest energy, E.

Energy, E is inversely proportional to the wavelength, λE, niλ


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