Flame Test: Analytical procedure to detect the presence of metal ions through flame color.
When heated, electrons in metal ions gain energy and transition to higher energy levels.
Electrons fall back to lower levels, releasing energy as light of characteristic colors.
Lithium (Li"): Red color
Sodium (Na"): Yellow color
Potassium (K+): Lilac color
Rubidium (Rb+): Red-Violet color
Caesium (Cs+): Blue color
Calcium (Ca²+): Orange color
Strontium (Sr²+): Red color
Barium (Ba²+): Green color
Radium (Ra²+): Intense color
Copper (Cu²+): Green/blue color
Iron (Fe²+/Fe³+): Yellow to brown color
Boron (B³+): Green color
Indium (In³+): Blue color
Lead (Pb²+): Blue color
Arsenic (As³+): Blue color
Antimony (Sb³+/Sb5+): Pale blue color
Selenium (Se²/Se⭑+): Red color
Zinc (Zn²+): Blue color
Note: Some metal ion colors are faint and hard to distinguish.
Exam 1: Scheduled for next Friday at 4:00 PM covering Modules 1 – 5 with 30 multiple-choice questions.
Calculator Requirement: Only approved syllabus calculators allowed.
Token Opportunity#1: Extended to Sunday night.
Token Opportunity#2: Open now – Complete the Chemical Naming video and practice assignment on Blackboard by Wednesday, February 19.
Wave Definition: Oscillation or periodic movement transporting energy.
Described by:
Wavelength (λ): Distance between two consecutive crests/troughs of a wave.
Frequency (ν): Number of wave cycles passing a stationary point per second (Hz).
Amplitude: Half distance between peaks and troughs.
Electromagnetic Spectrum: All types of electromagnetic radiation.
Speed of light (c): Constant speed for electromagnetic waves in a vacuum.
Relationship: 𝑐 = 𝜆ν
Where 𝑐 = 2.998 × 10^8 m/s.
Photon: Quantum of electromagnetic radiation, representing a "chunk" of energy.
Energy and frequency relationship: 𝑬 = 𝒉ν
h: Plank's constant = 6.626 × 10⁻³⁴ J·s.
Energy calculated for one photon:
𝑐 = 2.998 × 10^8 m/s
𝑬 = ℎ𝜈
𝜈 = 𝑐/𝜆
h = 6.63 × 10⁻³⁴ J·s.
Find frequency of electromagnetic radiation with wavelength 530.0 nm:
Use formula: 𝜈 = 𝑐/𝜆
Calculation: 𝜈 = 2.998 × 10^8 m/s / 5.30 × 10⁻⁷ m = 5.66 × 10¹⁴ s⁻¹
Calculate energy of one mole of photons of red light (632.8 nm):
Use formulas: 𝑬 = ℎν and 𝐸 = (6.63 × 10⁻³⁴ J·s) (3.0 × 10^8 m/s) / 6.328 × 10⁻⁷ m
Result: 1.89 × 10² kJ.
Light directed at metal surface can emit electrons if it has enough energy.
Demonstrates wave-particle duality of light.
Equation: 𝐸 = ℎν = ℎ𝑐/𝜆.
Continuous Spectrum: Unbroken series of wavelengths.
Line Spectrum: Narrow lines throughout spectral regions.
Each emission line corresponds to a single wavelength of light, indicating discrete energies of light emitted by gases.
Overview of atomic theory development:
John Dalton (1803): Indivisible atoms; different elements have varying atoms.
J.J. Thomson (1904): Electrons discovered; "Plum Pudding" model.
Ernest Rutherford (1911): Nucleus discovery through gold foil experiment.
Niels Bohr (1913): Electrons in quantized orbits; proposed modifications.
Electrons orbit nucleus in specific circular orbits with allowed energies.
The hydrogen atom does not emit energy while in fixed orbits.
Transition of electrons occurs via photon emission/absorption equal to energy difference between orbits.
Ground state (n=1): Lowest energy.
Energy level equation: 𝐸𝑛 = -𝑘/n².
As n increases, energy and distance from nucleus increase; energy levels get closer while orbits expand.
Excited state: Higher energy level absorption.
Photons emitted during electron transitions from higher to lower energy levels.
Conditions: Photon absorbed when ni < nf, emitted when ni > nf.
Emission lines correspond to transitions from excited states down to lower states.
Absorbed energy: ΔE positive; emitted energy: ΔE negative.
Photon energy = magnitude of ΔE.
Calculate frequency/wavelength using E = hν and λν = c.
Each element has a unique line spectrum.
Contributions:
Energy of electrons quantized.
Ability to transition between levels.
Shortcomings:
Only explains hydrogen's emission spectrum.
Circular orbits don't account for electron wave properties.
Find energy of a photon with a wavelength of 361 nm:
Calculation resulting in energy of 5.51 × 10⁻¹⁹ J.
Photons emitted when electron falls to lower energy levels.
Ground state transitions viable when a photon absorbs energy to a higher state.
Calculate frequency of photon with energy 1.93 × 10⁻¹⁷ J:
Result: 2.91 × 10¹⁶ s⁻¹.