1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-1 Chapter 7 Quantum Theory and Atomic Structure

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-2 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. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-3 Figure 7.1 Frequency and Wavelength c =  The Wave Nature ofLight

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-4 Figure 7.2 Amplitude (intensity) of a wave.

5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-5 Figure 7.3 Regions of the electromagnetic spectrum.

6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-6 Sample Problem 7.1 SOLUTION:PLAN: Interconverting Wavelength and Frequency wavelength in units given wavelength in m frequency (s -1 or Hz) = c/ Use c = 10 -2 m 1 cm 10 -9 m 1 nm = 1.00x10 -10 m = 325x10 -2 m = 473x10 -9 m == 3x10 8 m/s 1.00x10 -10 m = 3x10 18 s -1 == == 3x10 8 m/s 325x10 -2 m = 9.23x10 7 s -1 3x10 8 m/s 473x10 -9 m = 6.34x10 14 s -1 PROBLEM: A dental hygienist uses x-rays ( = 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky ( = 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, 3.00x10 8 m/s.) o 1 A = 10 -10 m 1 cm = 10 -2 m 1 nm = 10 -9 m o 325 cm 473nm 1.00A o 10 -10 m 1A o

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-7 Figure 7.4 Different behaviors of waves and particles.

8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-8 Figure 7.5 The diffraction pattern caused by light passing through two adjacent slits.

9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-9 Figure 7.6 Blackbody radiation E = n h  E =  n h  E = h when n = 1 smoldering coal electric heating elementlight bulb filament

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-10 Figure 7.7 Demonstration of the photoelectric effect.

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-11 Sample Problem 7.2 SOLUTION: PLAN: 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 1.20cm. What is the energy of one photon of this microwave radiation? After converting cm to m, we can use the energy equation, E = h combined with = c/ to find the energy. E = hc/ E = 6.626X10 -34 J*s3x10 8 m/s 1.20cm 10 -2 m cm x = 1.66x10 -23 J

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-12 Figure 7.8 The line spectra of several elements.

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-13 = RRydberg equation - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.096776x10 7 m -1 Figure 7.9 Three series of spectral lines of atomic hydrogen. for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-14 Figure 7.10 Quantum staircase.

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-15 Figure 7.11 The Bohr explanation of the three series of spectral lines.

16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-16 Figure B7.2 Figure B7.1 Emission and absorption spectra of sodium atoms. Flame tests. strontium 38 Srcopper 29 Cu

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-17

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-18 Figure B7.3 The main components of a typical spectrophotometer. Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected. Sample in compartment absorbs characteristic amount of each incoming wavelength. Computer converts signal into displayed data. Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths. Lenses/slits/collimaters narrow and align beam. Detector converts transmitted radiation into amplified electrical signal.

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-19 Figure 7.13 Wave motion in restricted systems.

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-20 Table 7.1 The de Broglie Wavelengths of Several Objects SubstanceMass (g)Speed (m/s) (m) slow electron fast electron alpha particle one-gram mass baseball Earth 9x10 -28 6.6x10 -24 1.0 142 6.0x10 27 1.0 5.9x10 6 1.5x10 7 0.01 25.0 3.0x10 4 7x10 -4 1x10 -10 7x10 -15 7x10 -29 2x10 -34 4x10 -63  h/m u

21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-21 Sample Problem 7.3 SOLUTION: PLAN: Calculating the de Broglie Wavelength of an Electron PROBLEM:Find the deBroglie wavelength of an electron with a speed of 1.00x10 6 m/s (electron mass = 9.11x10 -31 kg; h = 6.626x10 -34 kg*m 2 /s). Knowing the mass and the speed of the electron allows to use the equation = h/m u to find the wavelength. = 6.626x10 -34 kg*m 2 /s 9.11x10 -31 kgx1.00x10 6 m/s = 7.27x10 -10 m

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-22 Figure 7.14 Comparing the diffraction patterns of x-rays and electrons. x-ray diffraction of aluminum foilelectron diffraction of aluminum foil

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-23 CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. ObservationTheory Planck: Energy is quantized; only certain values allowed blackbody radiation Einstein: Light has particulate behavior (photons)photoelectric effect Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. atomic line spectra Figure 7.15 Summary of the major observations and theories leading from classical theory to quantum theory.

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-24 Since energy is wavelike perhaps matter is wavelike ObservationTheory deBroglie: All matter travels in waves; energy of atom is quantized due to wave motion of electrons Davisson/Germer: electron diffraction by metal crystal Since matter has mass perhaps energy has mass ObservationTheory Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. Compton: photon wavelength increases (momentum decreases) after colliding with electron Figure 7.15 continued QUANTUM THEORY Energy same as Matter particulate, massive, wavelike

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-25 The Heisenberg Uncertainty Principle  x * m  u ≥ h 44

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-26 Sample Problem 7.4 SOLUTION: PLAN: Applying the Uncertainty Principle PROBLEM:An electron moving near an atomic nucleus has a speed 6x10 6 ± 1% m/s. What is the uncertainty in its position (  x )? The uncertainty (  x ) is given as ±1%(0.01) of 6x10 6 m/s. Once we calculate this, plug it into the uncertainty equation.  u = (0.01)(6x10 6 m/s) = 6x10 4 m/s  x * m  u ≥ h 44 x ≥x ≥ 4  (9.11x10 -31 kg)(6x10 4 m/s) 6.626x10 -34 kg*m 2 /s = 9.52x10 -9 m

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-27 The Schrödinger Equation H  = E  d2d2 dy 2 d2d2 dx 2 d2d2 dz 2 ++ 82m82m h2h2 (E-V(x,y,z)  (x,y,z) = 0 + how  changes in space mass of electron total quantized energy of the atomic system potential energy at x,y,zwave function

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-28 Figure 7.16 Electron probability in the ground-state H atom.

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-29 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer 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

30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-30 Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed ValuesQuantum Numbers Principal, n (size, energy) Angular momentum, l (shape) Magnetic, m l (orientation) Positive integer (1, 2, 3,...) 0 to n-1 - l,…,0,…,+ l 1 0 0 2 01 0 3 0 12 0 0 +1 0+1 0 +2-2

31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-31 Sample Problem 7.5 SOLUTION: PLAN: 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 are allowed for n = 3? 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. For n = 3, l = 0, 1, 2 For l = 0 m l = 0 For l = 1 m l = -1, 0, or +1 For l = 2 m l = -2, -1, 0, +1, or +2 There are 9 m l values and therefore 9 orbitals with n = 3.

32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-32 Sample Problem 7.6 SOLUTION: PLAN: 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 Combine the n value and l designation to name the sublevel. Knowing l, we can find m l and the number of orbitals. n l sublevel namepossible m l values# of orbitals (a) (b) (c) (d) 3 2 5 4 2 0 1 3 3d 2s 5p 4f -2, -1, 0, 1, 2 0 -1, 0, 1 -3, -2, -1, 0, 1, 2, 3 5 1 3 7

33. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-33 Figure 7.17 1s2s3s

34. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-34 Figure 7.18 The 2p orbitals.

35. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-35 Figure 7.19 The 3d orbitals.

36. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-36 Figure 7.19 continued

37. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-37 Figure 7.20 One of the seven possible 4f orbitals.