1. 7-1 Dr. Wolf’s CHM 101 Chapter 7 Quantum Theory and Atomic Structure

2. 7-2 Dr. Wolf’s CHM 101 For Chapter 7: Be able to describe each of the following: Properties of light Photoelectric effect and blackbody radiation History of atomic theory in videos and book Bohr model and its significance Quantum mechanical model of the atom Be able to basically describe each orbital shape Qualitatively be able to describe each. What do they give us? Why are they important?: Rydberg Equation de Broglie wavelength Schrodinger equation Heisenberg Uncertainty Principle

3. 7-3 Dr. Wolf’s CHM 101 Wavelength,  the distance from one crest to the next in the wave. Measured in units of distance. c =  Frequency,  the number of complete cycles per sec., cps, Hz Electromagnetic Radiation ( light) - Wave like Speed of Light, C, same for all EM radiation. 3.00 x 10 8 m/sec in vacuum

4. 7-4 Dr. Wolf’s CHM 101 Mechanical waves: Energy is a function of the amplitude of wave

5. 7-5 Dr. Wolf’s CHM 101 Regions of the Electromagnetic Spectrum c =   = c / 1000 kHz100 MHz 3 m300 m In emf waves, energy is a function of the frequency

6. 7-6 Dr. Wolf’s CHM 101 Amplitude (Intensity) of a Wave

7. 7-7 Dr. Wolf’s CHM 101 Sample Problem 7.1 SOLUTION:PLAN: Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses x-rays ( = 1.00A) to take a series of dental radiographs while the patient listens to a radio station ( = 325cm) and looks out the window at the blue sky ( = 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, 3.00x10 8 m/s.) wavelength in units given wavelength in m frequency (s -1 or Hz) 1A = 10 -10 m 1cm = 10 -2 m 1nm = 10 -9 m = c/ Use c = 1.00A 325cm 473nm 10 -10 m 1A 10 -2 m 1cm 10 -9 m 1nm = 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

8. 7-8 Dr. Wolf’s CHM 101 Electromagnetic Radiation - Particle like 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.

9. 7-9 Dr. Wolf’s CHM 101 Demonstration of the photoelectric effect

10. 7-10 Dr. Wolf’s CHM 101 Electromagnetic Radiation - Particle like 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 = nh  where n is a positive integer, 1, 2, 3, etc. This means energy is quantized. E photon = h  =  E atom

11. 7-11 Dr. Wolf’s CHM 101 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. 7-12 Dr. Wolf’s CHM 101 Electromagnetic Radiation - Particle like 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.

13. 7-13 Dr. Wolf’s CHM 101 The line spectra of several elements

14. 7-14 Dr. Wolf’s CHM 101 = RRydberg equation - 1 1 n22n22 1 n12n12 R is the Rydberg constant = 1.096776 x 10 7 m -1 for the visible series, n 1 = 2 and n 2 = 3, 4, 5,... Three Series of Spectral Lines of Atomic Hydrogen Looking for an equation that would predict the wavelength seen in H spectrum But WHY does this equation work?

15. 7-15 Dr. Wolf’s CHM 101 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.

16. 7-16 Dr. Wolf’s CHM 101 Quantum staircase

17. 7-17 Dr. Wolf’s CHM 101 The Bohr explanation of the three series of spectral lines.

18. 7-18 Dr. Wolf’s CHM 101 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. But why must the electron’s energy be quantized?

19. 7-19 Dr. Wolf’s CHM 101 The de Broglie Wavelengths of Several Objects SubstanceMass (g)Speed, u, (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 E = hc/  h/m u so = hc/ E but E = m u c de Broglie Wavelength - giving particles wave-like properties

20. 7-20 Dr. Wolf’s CHM 101 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

21. 7-21 Dr. Wolf’s CHM 101 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 Summary of the major observations and theories leading from classical theory to quantum theory.

22. 7-22 Dr. Wolf’s CHM 101 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 QUANTUM THEORY Energy same as Matter particulate, massive, wavelike

23. 7-23 Dr. Wolf’s CHM 101 The Heisenberg Uncertainty Principle  x m  u h 44 Heisenberg Uncertainty Principle expresses a limitation on accuracy of simultaneous measurement of observables such as the position and the momentum of a particle.. 

24. 7-24 Dr. Wolf’s CHM 101 Sample Problem 7.4 SOLUTION: PLAN: Applying the Uncertainty Principle PROBLEM:An electron moving near an atomic nucleus has a speed 6 x 10 6 ± 1% m/s. What is the uncertainty in its position (  x )? The uncertainty (  u ) is given as ±1% (0.01) of 6 x 10 6 m/s. Once we calculate this, plug it into the uncertainty equation.  u = (0.01) (6 x 10 6 m/s) = 6 x 10 4 m/s  x m  u  h 44  x  4  (9.11 x 10 -31 kg) (6 x10 4 m/s) 6.626 x 10 -34 kg-m 2 /s  10 -9 m.

25. 7-25 Dr. Wolf’s CHM 101 The Schrödinger Equation - Quantum Numbers and Atomic Orbitals n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 S=0 p=1 d=2 f=3 m l the magnetic moment quantum number - an integer from - l to + l s has 1 p has 3 d has 5 f has 7 m s the spin quantum number, + 1/2 or - 1/2 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.

26. 7-26 Dr. Wolf’s CHM 101 Electron probability in the ground-state H atom “Orbital” showing 90% of electron probability n = 1 l = 0 m = 0

27. 7-27 Dr. Wolf’s CHM 101 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.

28. 7-28 Dr. Wolf’s CHM 101 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. l sublevel namepossible m l values# of orbitals (a) (b) (c) (d) 3d 2s 5p 4f -2, -1, 0, 1, 2 0 -1, 0, 1 -3, -2, -1, 0, 1, 2, 3 n 3 2 2 0 5 1 4 3 5 1 3 7

29. 7-29 Dr. Wolf’s CHM 101 Table 8.1 Summary of Quantum Numbers of Electrons in Atoms NameSymbolPermitted ValuesProperty principalnpositive integers (1, 2, 3, …)orbital energy (size) angular momentum l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magneticmlml integers from - l to 0 to + l orbital orientation spinmsms +½ or -½direction of e - spin

30. 7-30 Dr. Wolf’s CHM 101

31. 7-31 Dr. Wolf’s CHM 101 s orbitals 1s2s3s

32. 7-32 Dr. Wolf’s CHM 101 The 2p orbitals n = 2, l = 1 p orbitals - three of them Combination

33. 7-33 Dr. Wolf’s CHM 101 The 3d orbitals n = 3 l = 2 d orbitals - five of them

34. 7-34 Dr. Wolf’s CHM 101 d orbitals - five of them Combination

35. 7-35 Dr. Wolf’s CHM 101 One of the seven possible 4f orbitals f orbitals - seven of them