Chapter 7 - Quantum Theory and the Electronic Structure of...

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Quantum Theory and the Electronic Structure of Atoms Chapter 7
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OK, so an atom consists of a nucleus containing positively-charged proton(s) and neutron(s) , around which resides constantly-moving and negatively-charged electron(s) , right? But classical physics dictates that a negatively-charged particle moving in a curved path around a positively-charged particle MUST emit radiation and thus lose energy continuously…. .so eventually, electron(s) should spiral into the nucleus. But they don’t. Why? This problem forced a complete re-thinking of the classical picture of matter and energy . The distinctions between the discrete particulate nature of matter and the continuous wave-like nature of energy were to start to fade……… wave-particulate duality of matter and energy . First, let’s take a look at electromagnetic radiation (radiant energy) . Then, we’ll see how the wave-particulate duality of matter and energy applies to the electronic structure of atoms.
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Properties of Waves Wavelength ( λ ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough.
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Properties of Waves Frequency ( ν ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed ( u ) of the wave = λ x ν
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Maxwell (1873), proposed that visible light consists of electromagnetic waves . Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light ( c ) in vacuum = 3.00 x 10 8 m/s All electromagnetic radiation λ x ν = c
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λ x ν = c λ = c/ ν λ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz λ = 5.0 x 10 3 m Radio wave A certain electromagnetic radiation has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? λ = 5.0 x 10 12 nm λ ν
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Mystery #1, “Black Body Problem” Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x ν Planck’s constant (h) h = 6.626 x 10 -34 J s Only certain “quanta” of energy are emitted or absorbed, or E = nhν where n = the quantum number (1,2,3…). So the ΔE = Δnhν and the smallest ΔE is when Δn=1. Also, E = hc/λ
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Light has both: 1. wave nature 2. particle nature Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light h ν KE e - E photon = hν = hc/λ (A mole of photons = 6.02x10 23 photons) (E = mc 2 )
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E = h x ν E = 6.63 x 10 -34 (J s) x 3.00 x 10 8 (m/s) / 0.154 x 10 -9 (m) E = 1.29 x 10 -15 J E = h x c / λ When copper is bombarded with high-energy photons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.
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Line Emission Spectrum of Hydrogen Atoms
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1. e - resides in certain allowable (quantized) energy levels 1. A photon (light) is emitted
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This note was uploaded on 07/15/2011 for the course CHEM 2045 taught by Professor Gower during the Spring '11 term at University of Florida.

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Chapter 7 - Quantum Theory and the Electronic Structure of...

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