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 Demonstration of the photoelectric effect.

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-10 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

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-11 Figure 7.7 The line spectra of several elements.

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-12 = RRydberg equation - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.096776x10 7 m -1 Figure 7.8 Three series of spectral lines of atomic hydrogen. for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-13 Figure 7.9 Quantum staircase.

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-14 Figure 7.10 The Bohr explanation of the three series of spectral lines.

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-15 Figure 7.13 Wave motion in restricted systems.

16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-16 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

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

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-18 Figure 7.14 Comparing the diffraction patterns of x-rays and electrons. x-ray diffraction of aluminum foilelectron diffraction of aluminum foil

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-19 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.

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-20 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

21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-21 The Heisenberg Uncertainty Principle  x * m  u ≥ h 44

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-22 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

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-23 Figure 7.16 Electron probability in the ground-state H atom.

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-24 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

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-25 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

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-26 Sample Problem 7.4 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.

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-27 Sample Problem 7.5 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

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-28 Figure 7.18

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-29 Figure 7.19 The 2p orbitals.

30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-30 Figure 7.20 The 3d orbitals.

31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-31 Figure 7.20 continued

32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 7-32 Figure 7.21 One of the seven possible 4f orbitals.