Chapter 7. Quantum Theory and Atomic Structure. Quantum Mechanics. Chap 7-1
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1 Chapter 7 Quantum Theory and Atomic Structure Chap 7-1
2 Quantum Theory and Atomic Structure 7.1 The Nature of Light 7.2 Atomic Spectra 7.3 The Wave-Particle Duality of Matter and Energy 7.4 The Quantum-Mechanical Model of the Atom Chap 7-2
3 Frequency and wavelength. c = λ ν c = x 10 8 m/sec Chap 7-3
4 Amplitude (intensity) of a wave. Chap 7-4
5 Regions of the electromagnetic spectrum. Chap 7-5
6 Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses x-rays (λ = 1.00 Ǻ) to take a series of dental radiographs while the patient listens to a radio station (λ = 325 cm) and looks out the window at the blue sky (λ = 473 nm). What is the frequency (in s -1 ) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00 x 10 8 m/s.) PLAN: Use c = λν Chap 7-6
7 Interconverting Wavelength and Frequency: Solution c = λν 1.00 Ǻ x m 1 Ǻ = 1.00 x m 3 x 10 8 m/s ν = 1.00 x m = 3 x s cm x 10-2 m 1 cm = 325 x 10-2 m ν = 3 x 10 8 m/s 325 x 10-2 m = 9.23 x 10 7 s nm x 10-9 m 1 nm = 473 x 10-9 m ν = 3 x 10 8 m/s 473 x 10-9 m = 6.34 x s -1 Chap 7-7
8 Different behaviors of waves and particles. Chap 7-8
9 The diffraction pattern caused by light passing through two adjacent slits. Chap 7-9
10 Particle Nature of Light: Blackbody Radiation Three phenomena could not be explained by classical physics: blackbody radiation, photoelectric effect, and atomic spectra. A new picture of energy was required. Blackbody Radiation: Heat a solid object: at 1000 K it begins to emit soft red light; at 1500 K it begins to glow orange and is brighter; at 2000 K it is still brighter and white. How to explain? Planck: proposed that only certain quantities of energy could be emitted or absorbed: E = nhν = nhc λ These packets of energy are called quanta, and n is a quantum number. Since it was (correctly) assumed that most transitions occur between adjacent states, Δn = 1 and the change in energy is: ΔE = hν = hc λ Chap 7-10
11 Demonstration of the photoelectric effect. NOT a function of light intensity; IS a function of light wavelength (and therefore frequency) Threshold frequency ; NO time lag. Chap 7-11
12 Calculating the Energy of Radiation from Its Wavelength PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20 cm. What is the energy of one photon of this microwave radiation? PLAN: After converting cm to m, we can use the energy equation, E = hν combined with ν = c/λ to find the energy. SOLUTION: E = hν = hc λ E = ( J sec)( m / sec) m = J Chap 7-12
13 Atomic Spectra: A Problem to Solve! Continuum : Classical Physics Chap 7-13
14 The line spectra of several elements. Chap 7-14
15 Rydberg equation for emission 1 λ = R 1 2 n 1 2 where n 2 > n 1 1 n 2 R is the Rydberg constant = x 10 7 m -1 Three series of spectral lines of atomic hydrogen. for the visible series, n 1 = 2 and n 2 = 3, 4, 5,... Chap 7-15
16 Quantum staircase. Chap 7-16
17 The Bohr explanation of the three series of spectral lines. Chap 7-17
18 Wave motion in restricted systems. Chap 7-18
19 Wave-Particle Duality Energy (electromagnetic radiation) behaves like waves (diffraction, wavelength, frequency) but also possesses particle behavior (photons in photoelectric effect). Invoke wave-particle duality of EMR to explain. If energy can exhibit both particle and wave natures, what about matter? debroglie: Matter Waves Large items have very small wavelengths Electrons are small enough however to exhibit measurable wave properties Chap 7-19
20 λ = h mv = h p Chap 7-20
21 Determining E and λ of an Electron Transition PROBLEM: A hydrogen atom absorbs a photon of visible light and its electron enters the n = 4 energy level. Calculate (a) the change in energy of the atom and (b) the wavelength (in nm) of the photon. PLAN: (a) The H atom absorbs visible light, so the electron is going from n = 2 to n = 4. Calculate ΔE. (b) ΔE = hν = hc/λ SOLUTION: (a) ΔE = J ΔE = J Z 2 n final n initial Z = J (b) λ = hc ΔE = ( J s)( m/s) J = m = 486 nm Chap 7-21
22 Comparing diffraction patterns of x-rays and electrons. X-ray diffraction of aluminum Electron diffraction of aluminum Chap 7-22
23 Summary of the major observations and theories leading from classical theory to quantum theory. Chap 7-23
24 Wave-Particle Duality: Problems Classical mechanics: particles have trajectories and can precisely be described according to location (x) and linear momentum (p) Quantum mechanics: can t think like this: waves are spread out as they oscillate Duality denies the ability of knowing the trajectory of particles (complementarity) Heisenberg Uncertainty principle: Important for wave equations Chap 7-24
25 The Heisenberg Uncertainty Principle Δ x m Δ u h 4π Chap 7-25
26 Schrödinger Equation The SWE relates the second derivative of Ψ to the value of Ψ at each point; impossible to solve exactly (except simple cases) One simple solution: particle in a box Solution of SWE using this yields: Λ = λενγτη οφ βοξ ν = 1,2,3 ξ = διστανχε βετωεεν 0 ανδ Λ Since n is an integer, E is quantized with discrete values of E and, since ΔE = hν and ΔE = E n+1 - E n Chap 7-26
27 SWE: Particle in a Rectangular What if the box is 3-dimensional? Box Solution of SWE using this yields: Chap 7-27
28 Electron probability density in the ground-state H atom. Chap 7-28
29 Hydrogen Wavefunctions Angle independent R decreases exponentially as r increases a 0 is the Bohr radius Chap 7-29
30 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer ( 1, 2, 3, ) l the angular momentum quantum number - an integer from 0 to (n - 1) m l the magnetic moment quantum number - an integer from l to +l Chap 7-30
31 Chap 7-31
32 Determining Quantum Numbers for an Energy Level PROBLEM: What values of the angular momentum (l) and magnetic (m l ) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals exist for n = 3? PLAN: Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to (n 1); m l can be integers from -l through 0 to +l. SOLUTION: For n = 3, l = 0, 1, 2 For l = 0, m l = 0 For l = 1, m l = -1, 0, +1 For l = 2, m l = -2, -1, 0, +1, +2 There are 9 m l values and therefore, 9 orbitals with n = 3. Chap 7-32
33 Determining Sublevel Names and Orbital Quantum Numbers PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: (a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3 PLAN: Combine the n value and l designation to name the sublevel. Knowing l, we can find m l and the number of orbitals. SOLUTION: n l sublevel name possible m l values # of orbitals (a) 3 2 3d -2, -1, 0, +1, +2 5 (b) 2 0 2s 0 1 (c) 5 1 5p -1, 0, +1 3 (d) 4 3 4f -3, -2, -1, 0, +1, +2, +3 7 Chap 7-33
34 The 1s, 2s, and 3s orbitals. Chap 7-34
35 The 2p orbitals. Chap 7-35
36 The 3d orbitals. Chap 7-36
37 One of the seven possible 4f orbitals. Chap 7-37
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