Ch.7 - Quantum MechanicsWorksheetSee all chapters
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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 value of n for which the wavelength of a Balmer series line is smaller than 400 nm, which is the lower limit for wavelengths in the visible spectrum?

Problem

What is the smallest value of n for which the wavelength of a Balmer series line is smaller than 400 nm, which is the lower limit for wavelengths in the visible spectrum?

Solution

We're asked to determine the smallest value of n (ni with a corresponding wavelength, λ that is smaller than 400 nm for a Balmer series line.  400 nm is the lower limit for wavelengths in the visible spectrum.

Recall the Balmer Equation shown below:


1λ=R×1nf2-1ni2

where:

λ  = wavelength, m 

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. 


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