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
Problem

If a single electron in a hydrogen atom is in the excited n = 4 state and eventually relaxes to lower states, there are six possible emission or spectral lines.  How many of these lines are in the visible region of the EM spectrum?

a.  1

b. 2

c. 3

d. 4

e. all of them

Solution

Calculate each emission state from n = 4 to n = 3, 2 and 1 using Balmer equation and find which will produce a wavelength at Visible spectrum

We’re going to use the Balmer Equation which relates wavelengths to a photon’s electronic transitions.

1λ=RZ21n2final-1n2initial

λ = wavelength, m
R = Rydberg constant = 1.097x107 m-1 
Z = atomic number of the element
ninitial = initial energy level
nfinal = final energy level


According to the electromagnetic spectrum, visible light has a wavelength at the range 400-700 nm or 4x10-7- 7x10-7 m

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