1. (b) = 0.18 cm In this case the wavelength is significant. While the De Broglie equation applies to all systems, the wave properties become observable only for microscopic objects. 541

2. An application of De Broglie’s idea 542

3. An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. 543

4. An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. So for visible light at 400 nm, you could see an object with a size of around 200 nm or 2x10 -5 cm. 544

5. An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. So for visible light at 400 nm, you could see an object with a size of around 200 nm or 2x10 -5 cm. To “see” something smaller, could use X-rays – but its hard to focus X-rays. 545

6. However, could work with electrons in place of light. 546

7. However, could work with electrons in place of light. Relatively easy to accelerate electrons to high velocity, so from the formula above, the wavelength would be small. 547

8. However, could work with electrons in place of light. Relatively easy to accelerate electrons to high velocity, so from the formula above, the wavelength would be small. Can now see individual atoms. 548

9. The Heisenberg Uncertainty Principle 549

10. The Heisenberg Uncertainty Principle In 1927 Heisenberg proposed a principle of great importance in the philosophical groundwork of quantum theory. 550

11. The Heisenberg Uncertainty Principle In 1927 Heisenberg proposed a principle of great importance in the philosophical groundwork of quantum theory. Heisenberg’s principle is primarily concerned with measurements on the atomic scale. 551

12. In classical physics it was generally believed that the accuracy with which any quantity could be determined depended upon the instrument used to do the measurements. 552

13. In classical physics it was generally believed that the accuracy with which any quantity could be determined depended upon the instrument used to do the measurements. Thought experiment: Suppose we wish to monitor the position and velocity of a tennis ball in flight. Can do this with different instruments. Important consideration – must be able to “see” the tennis ball – simply shine light on the ball. 553

14. Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. 554

15. Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: 555

16. Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: E = h and hence. 556

17. Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: E = h and hence. Therefore the light has high energy if the wavelength is very small. 557

18. The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. 558

19. The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. Thus at the very instant we determine the position of the electron – we lose information on its momentum. 559

20. The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. Thus at the very instant we determine the position of the electron – we lose information on its momentum. To reduce the impact we could choose light of a longer wavelength (i.e. lower energy), but then we would be unable to pinpoint the position of the electron. 560

21. This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: 561

22. This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: It is impossible to determine to arbitrary accuracy both the position and momentum of a particle at the same time. 562

23. This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: It is impossible to determine to arbitrary accuracy both the position and momentum of a particle at the same time. Let x represent the position and p the momentum of a particle. Let and denote the uncertainties in the measurements (i.e. the measurement errors), then 563

24. Heisenberg showed that 564

25. Heisenberg showed that Of course it always possible to do a sloppy experiment, then the product of the two uncertainties could be a lot larger than. 565

26. Heisenberg showed that Of course it always possible to do a sloppy experiment, then the product of the two uncertainties could be a lot larger than. Like the De Broglie relation, the Heisenberg Uncertainty Principle applies to both macroscopic and microscopic objects. 566

27. The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. 567

28. The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. Consequently, it is impossible to pin down the exact nature of the electron. 568

29. The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. Consequently, it is impossible to pin down the exact nature of the electron. Whether an electron should be thought of as a particle, a wave, or a particle-wave – depends on how we take a measurement. 569

30. When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. 570

31. When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. When we speak of wavelength of an electron – we are talking about the electron as a wave. 571

32. When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. When we speak of wavelength of an electron – we are talking about the electron as a wave. There are some additional properties of the electron that we will encounter that will tie in directly with the wave-like nature of the electron. 572

33. Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x 10 -6 cm s -1. Calculate the corresponding uncertainty in determining its position. 573

34. Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x 10 -6 cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v 574

35. Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x 10 -6 cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v so = m 575

36. Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x 10 -6 cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v so = m = (9. 11 x 10 -28 g)(1.0 x 10 -6 cm s -1 ) = 9.1 x 10 -34 g cm s -1 576

37. From, for the purpose of calculation, we can treat the as an = sign. Hence, 577

38. From, for the purpose of calculation, we can treat the as an = sign. Hence, 578

39. From, for the purpose of calculation, we can treat the as an = sign. Hence, = 5.8 x 10 3 m (or 3.6 miles!) 579

40. Exercise: Try a similar calculation on your instructor (or yourself)! Assume a mass of around 80 kg (or your mass) and an uncertainty in velocity of around 0.5 cm s -1. Is the uncertainty in the instructor’s position (your position) so uncertain that he (you) is (are) not in the room? 580

41. Quantum Mechanics The spectacular initial success of Bohr’s theory was followed by a series of disappointments. 581

42. Quantum Mechanics The spectacular initial success of Bohr’s theory was followed by a series of disappointments. Bohr (nor anyone else) could account for the emission spectra of atoms more complex than the H atom. 582

43. It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. 583

44. It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. The key figure to make the great advance was Schrödinger (1926). 584

45. It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. The key figure to make the great advance was Schrödinger (1926). Schrödinger formulated a new equation to describe the motion of microscopic bodies. 585

46. Schrödinger formulated an equation that incorporated the particle properties in terms of the mass (m) and charge (e) and the wave properties in terms of a function – which is called the wave function. The symbol employed for the wave function is (psi). 586

47. Schrödinger formulated an equation that incorporated the particle properties in terms of the mass (m) and charge (e) and the wave properties in terms of a function – which is called the wave function. The symbol employed for the wave function is (psi). There is no simple physical meaning for the function. However the function has the meaning that it gives the probability of finding the particle at a particular point in space. 587

48. This probability thinking is a radical break with classical thinking. 588

49. The Hydrogen atom and the Schr ö dinger equation 589

50. The Hydrogen atom and the Schr ö dinger equation One of the first problems investigated by Schrödinger was the energy levels for the H atom. 590

51. The Hydrogen atom and the Schr ö dinger equation One of the first problems investigated by Schrödinger was the energy levels for the H atom. Schrödinger obtained the result: 591

52. 592

53. where: m = the mass of the electron 593

54. where: m = the mass of the electron e = charge on the electron 594

55. where: m = the mass of the electron e = charge on the electron h = Planck’s constant 595

56. where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number 596

57. where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number = vacuum permittivity 597

58. where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number = vacuum permittivity E n = the energy of the different states 598

59. This formula shows that the energies of an electron in the hydrogen atom are quantized. 599

60. This formula shows that the energies of an electron in the hydrogen atom are quantized. The above formula will be abbreviated as k E n = – n 2 600

61. It should be no surprise that the Schrödinger result exactly matches the Bohr result. Recall that Bohr was able to explain the spectrum of the H atom, exactly matching the best experimental data available at the time. 601

62. Explanation of the spectrum of the H atom 602

63. Explanation of the spectrum of the H atom Energy Level Diagram for the H atom 603

64. Explanation of the spectrum of the H atom Energy Level Diagram for the H atom Consider two energy levels for the H atom, the upper one characterized by the quantum number n upper (abbreviated n u ) and a lower level characterized by the quantum number n lower (abbreviated n l ). 604

65. Two situations: (1) absorption and (2) emission. n u n l absorption (photon absorbed) 605

66. n u n l emission (photon emitted) 606

67. In both the absorption and emission processes Recall so that 607

68. In both the absorption and emission processes Recall so that k From the formula E n = - we can n 2 write : 608

69. and 609

70. and Since we have 610

71. and Since we have 611

72. and Since we have 612

73. and Since we have 613

74. That is 614

75. That is Set 615

76. That is Set So that 616

77. That is Set So that This is called the Rydberg formula and R H is called the Rydberg constant. 617

78. Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. 618

79. Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = 109678 cm -1 and k = 2.17869 x 10 -18 J 619

80. Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = 109678 cm -1 and k = 2.17869 x 10 -18 J Approach 1 620

81. Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = 109678 cm -1 and k = 2.17869 x 10 -18 J Approach 1 621

82. Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = 109678 cm -1 and k = 2.17869 x 10 -18 J Approach 1 622

83. = 23032.4 cm -1 623

84. = 23032.4 cm -1 Hence = 4.34171 x10 -5 cm = 434.171 nm 624

85. = 23032.4 cm -1 Hence = 4.34171 x10 -5 cm = 434.171 nm Now use 625

86. = 4.57528 x 10 -19 J 626

87. = 4.57528 x 10 -19 J Approach 2 Use k E n = - n 2 627

88. So that k k E 5 = - E 2 = - 5 2 2 2 628

89. So that k k E 5 = - E 2 = - 5 2 2 2 = -8.71476 x 10 -20 J = -5.44673 x 10 -19 J 629

90. So that k k E 5 = - E 2 = - 5 2 2 2 = -8.71476 x 10 -20 J = -5.44673 x 10 -19 J Therefore = E 5 - E 2 = 4.57525 x 10 -19 J 630