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Chapter 7 Quantum Theory and Atomic Structure

Chapter 7 Quantum Theory and Atomic Structure. 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. Electromagnetic Radiation ( light) - Wave like.

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Chapter 7 Quantum Theory and Atomic Structure

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  1. Chapter 7 Quantum Theory and Atomic Structure

  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

  3. Electromagnetic Radiation (light) - Wave like Wavelength, l, the distance from one crest to the next in the wave. Measured in units of distance. Frequency, n, the number of complete cycles per sec., cps, Hz Speed of Light, C, same for all EM radiation.3.00 x 1010 cm/sec in vacuum c = l n

  4. Amplitude (Intensity) of a Wave

  5. Regions of the Electromagnetic Spectrum c = l n n = c / l 3 m 300 m 100 MHz 1000 kHz

  6. PROBLEM: A dental hygienist uses x-rays (l= 1.00A) to take a series of dental radiographs while the patient listens to a radio station (l = 325cm) and looks out the window at the blue sky (l= 473nm). What is the frequency (in s-1) of the electromagnetic radiation from each source? (Assume that the radiation travels at the speed of light, 3.00x108m/s.) 1.00A 10-10m 1A 1A = 10-10m 1cm = 10-2m 1nm = 10-9m 325cm 10-2m 1cm n = c/l 10-9m 473nm 1nm 3x108m/s 473x10-9m Sample Problem 7.1 Interconverting Wavelength and Frequency PLAN: Use c = ln SOLUTION: = 1.00x10-10m wavelength in units given 3x108m/s = 3x1018s-1 n = 1.00x10-10m = 325x10-2m 3x108m/s wavelength in m n = = 9.23x107s-1 325x10-2m = 473x10-9m frequency (s-1 or Hz) = 6.34x1014s-1 n =

  7. Different behaviors of waves and particles.

  8. The diffraction pattern caused by light passing through two adjacent slits.

  9. Electromagnetic Radiation - Particlelike The view that EM was wavelike could not explain certain phenomena like : 1) Blackbody radiation - when objects are heated, they give off shorter and more intense radiation as the temperature increases, e.g. dull red hot, to hotter orange, to white hot....... But wavelike properties would predict hotter temps would continue to give more and more of shorter and shorter wavelengths. But instead a bell shape curve of intensities is obtained with the peak at the IR- Vis part of the spectrum.

  10. Electromagnetic Radiation E = hn E = hc/l Blackbody Radiation Planck’s constanth = 6.626 x 10-34 J-s

  11. Electromagnetic Radiation - Particlelike The view that EM was wavelike could not explain certain phenomena like : 2) Photoelectron Effect - light shinning on certain metal plates caused a flow of electrons. However the the light had a minimum frequency to cause the effect, i.e. not any color would work. And although bright light caused more electron flow than weak light, electron flow started immediately with both strong or weak light.

  12. Demonstration of the photoelectric effect

  13. Electromagnetic Radiation - Particlelike The better explanation for these experiments was that EM consisted of packets of energy called photons (particle-like) that had wave-like properties as well. And that atoms could have only certain quantities of energy, E = nhn , where n is a positive integer, 1, 2, 3, etc. This means energy is quantized. Ephoton = hn = Eatom

  14. PROBLEM: A cook uses a microwave oven to heat a meal. The wavelength of the radiation is 1.20cm. What is the energy of one photon of this microwave radiation? 10-2m cm Sample Problem 7.2 Calculating the Energy of Radiation from Its Wavelength PLAN: After converting cm to m, we can use the energy equation, E = hn combined with n = c/l to find the energy. SOLUTION: E = hc/l 6.626X10-34J-s x 3x108m/s E = = 1.66x10-23J 1.20cm

  15. Electromagnetic Radiation - Particlelike The view that EM was wavelike could not explain certain phenomena like : 3) Atomic Spectra - Electrical discharges in tube of gaseous elements produces light (EM).But not all wavelengths of light were produced but just a few certain wavelengths (or frequencies). And different elements had different wavelengths associated with them. Not just in the Visible but also IR and UV regions.

  16. The line spectra of several elements

  17. 1 1 1 l n12 n22 Three Series of Spectral Lines of Atomic Hydrogen Looking for an equation that would predict the wavelength seen in H spectrum Rydberg equation = R - for the visible series, n1 = 2 and n2 = 3, 4, 5, ... R is the Rydberg constant = 1.096776 x 107 m-1 But WHY does this equation work?

  18. Bohr Model of the Hydrogen Atom Assumed the H atom has only certain allowable energy levels for the electron orbits. (quantized because it made the equations work)) When the electron moves from one orbit to another, it has to absorb or emit a photon whose energy equals the difference in energy between the two orbits. By assuming the electron traveled in circular orbits, the energy level for each orbits was E = -2.18 x 10-18 J / n2 where n = 1, 2, 3,.... Note: because of negative sign, lowest energy (most stable) when n = 1 and highest energy is E = 0 when n = . So energy released when the electron moves from one n level to another is E= hn = hc/l = -2.18 x 10-18 J ( 1 / n2final - 1 / n2initial )

  19. Quantum staircase

  20. The Bohr explanation of the three series of spectral lines.

  21. But why must the electron’s energy be quantized? If EM can have particle-like properties in addition to being wave-like, what if the electron particles have wave-like properties? Quantization is a natural consequence of having wave-like properties.

  22. Wave motion in restricted systems

  23. h /mu l = de Broglie Wavelength -giving particles wave-like properties but E = muc E = hc/l so l = hc/ E The de Broglie Wavelengths of Several Objects Substance Mass (g) Speed, u, (m/s) l (m) slow electron 9x10-28 1.0 7x10-4 fast electron 9x10-28 5.9x106 1x10-10 alpha particle 6.6x10-24 1.5x107 7x10-15 one-gram mass 1.0 0.01 7x10-29 baseball 142 25.0 2x10-34 Earth 6.0x1027 3.0x104 4x10-63

  24. PROBLEM: Find the deBroglie wavelength of an electron with a speed of 1.00x106m/s (electron mass = 9.11x10-31kg; h = 6.626x10-34 kg-m2/s). Sample Problem 7.3 Calculating the de Broglie Wavelength of an Electron PLAN: Knowing the mass and the speed of the electron allows to use the equation l = h/mu to find the wavelength. SOLUTION: 6.626x10-34kg-m2/s l = = 7.27x10-10m 9.11x10-31kg x 1.00x106m/s

  25. Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. blackbody radiation Planck: Energy is quantized; only certain values allowed photoelectric effect Einstein: Light has particulate behavior (photons) atomic line spectra Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Summary of the major observations and theories leading from classical theory to quantum theory. Observation Theory

  26. Since energy is wavelike, perhaps matter is wavelike Davisson/Germer: electron diffraction by metal crystal deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons Since matter has mass, perhaps energy has mass Observation Theory Compton: photon wavelength increases (momentum decreases) after colliding with electron Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. QUANTUM THEORY Energy same as Matter particulate, massive, wavelike Observation Theory

  27. h 4p The Heisenberg Uncertainty Principle Heisenberg Uncertainty Principleexpresses a limitation on accuracy of simultaneous measurement of observables such as the position and the momentum of a particle. . D xmD u 

  28. PROBLEM: An electron moving near an atomic nucleus has a speed 6 x 106 ± 1% m/s. What is the uncertainty in its position (D x)? h D xmD u 4p Sample Problem 7.4 Applying the Uncertainty Principle PLAN: The uncertainty (D u) is given as ±1% (0.01) of 6 x 106 m/s. Once we calculate this, plug it into the uncertainty equation. SOLUTION: D u = (0.01) (6 x 106 m/s) = 6 x 104 m/s . 6.626 x 10-34 kg-m2/s D x  10-9m 4p (9.11 x 10-31 kg) (6 x104 m/s)

  29. The Schrödinger Equation - Quantum Numbers and Atomic Orbitals A complicated equation with multiple solutions which describes the probability of locating an electron at the various allowed energy levels. Solutions involve three interdependent variables to describe an electron orbital. i.e., an atomic orbital is specified by three quantum numbers. nthe principal quantum number - a positive integer lthe angular momentum quantum number - an integer from 0 to n-1 mlthe magnetic moment quantum number - an integer from -l to +l msthe spin quantum number, + 1/2 or - 1/2

  30. Electron probability in the ground-state H atom “Orbital” showing 90% of electron probability n = 1 l = 0 m = 0

  31. 0 1 2 0 -1 0 +1 -1 0 +1 -2 -1 0 +1 +2 Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum Numbers Principal, n (size, energy) Positive integer (1, 2, 3, ...) 1 2 3 Angular momentum, l (shape) 0 to n-1 0 0 1 sublevel namesl = 0, called “s” l = 1, “p”l = 2, “d” l = 3, “f” 0 0 Magnetic, ml (orientation) -l,…,0,…,+l

  32. PROBLEM: What values of the angular momentum (l) and magnetic (ml) quantum numbers are allowed for a principal quantum number (n) of 3? How many orbitals are allowed for n = 3? Sample Problem 7.5 Determining Quantum Numbers for an Energy Level PLAN: Follow the rules for allowable quantum numbers found in the text. l values can be integers from 0 to n-1; ml can be integers from -l through 0 to + l. SOLUTION: For n = 3, l = 0, 1, 2 For l = 0 ml = 0 For l = 1 ml = -1, 0, or +1 For l = 2 ml = -2, -1, 0, +1, or +2 There are 9 mlvalues and therefore 9 orbitals with n = 3.

  33. PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: Sample Problem 7.6 Determining Sublevel Names and Orbital 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 ml and the number of orbitals. SOLUTION: n l sublevel name possible ml 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

  34. s orbitals 1s 2s 3s

  35. p orbitals - three of them Combination The 2p orbitals n = 2, l = 1

  36. d orbitals - five of them The 3d orbitals n = 3 l = 2

  37. d orbitals - five of them Combination

  38. f orbitals - seven of them One of the seven possible 4f orbitals

  39. End of Chapter 7

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