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Chapter 7 Quantum Theory and Atomic Structure 7-1 Dr. Wolf’s CHM 101 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 7-2 Dr. Wolf’s CHM 101 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=ln 7-3 Dr. Wolf’s CHM 101 Amplitude (Intensity) of a Wave 7-4 Dr. Wolf’s CHM 101 c=ln Regions of the Electromagnetic Spectrum n =c/l 3m 100 MHz 7-5 Dr. Wolf’s CHM 101 300 m 1000 kHz Sample Problem 7.1 Interconverting Wavelength and Frequency 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.) PLAN: Use c = ln wavelength in units given 1A = 10-10m 1cm = 10-2m 1nm = 10-9m wavelength in m n = c/l frequency (s-1 or Hz) 7-6 Dr. Wolf’s CHM 101 SOLUTION: 1.00A 10-10m = 1.00x10-10m 1A 3x108m/s = 3x1018s-1 n= 1.00x10-10m -2 325cm 10 m = 325x10-2m 1cm 3x108m/s n= = 9.23x107s-1 325x10-2m 10-9m 473nm = 473x10-9m 1nm 3x108m/s = 6.34x1014s-1 n= 473x10-9m Different behaviors of waves and particles. 7-7 Dr. Wolf’s CHM 101 The diffraction pattern caused by light passing through two adjacent slits. 7-8 Dr. Wolf’s CHM 101 Electromagnetic Radiation - Particle like 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. 7-9 Dr. Wolf’s CHM 101 Electromagnetic Radiation E = hn Blackbody Radiation E = hc/l Planck’s constant h = 6.626 x 10-34 J-s 7-10 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. 7-11 Dr. Wolf’s CHM 101 Demonstration of the photoelectric effect 7-12 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 = nhn , where n is a positive integer, 1, 2, 3, etc. This means energy is quantized. Ephoton = hn = Eatom 7-13 Dr. Wolf’s CHM 101 Sample Problem 7.2 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? 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= E = hc/l 6.626X10-34J-s x 3x108m/s 1.20cm 10-2m cm 7-14 Dr. Wolf’s CHM 101 = 1.66x10-23J 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. 7-15 Dr. Wolf’s CHM 101 The line spectra of several elements 7-16 Dr. Wolf’s CHM 101 Three Series of Spectral Lines of Atomic Hydrogen Looking for an equation that would predict the wavelength seen in H spectrum Rydberg equation 1 l = R is the Rydberg constant = 1.096776 x R 107 1 2 n1 m-1 But WHY does this equation work? 7-17 Dr. Wolf’s CHM 101 1 n22 for the visible series, n1 = 2 and n2 = 3, 4, 5, ... 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 ) 7-18 Dr. Wolf’s CHM 101 Quantum staircase 7-19 Dr. Wolf’s CHM 101 The Bohr explanation of the three series of spectral lines. 7-20 Dr. Wolf’s CHM 101 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. 7-21 Dr. Wolf’s CHM 101 Wave motion in restricted systems 7-22 Dr. Wolf’s CHM 101 de Broglie Wavelength E = hc/l giving particles wave-like properties so l = hc/ E but E = muc l = h /mu The de Broglie Wavelengths of Several Objects Substance Speed, u, (m/s) l (m) slow electron 9x10-28 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 6.0x1027 3.0x104 4x10-63 Earth 7-23 Mass (g) Dr. Wolf’s CHM 101 1.0 7x10-4 Sample Problem 7.3 PROBLEM: PLAN: Calculating the de Broglie Wavelength of an Electron 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). Knowing the mass and the speed of the electron allows to use the equation l = h/mu to find the wavelength. SOLUTION: l= 6.626x10-34kg-m2/s 9.11x10-31kg x 1.00x106m/s 7-24 Dr. Wolf’s CHM 101 = 7.27x10-10m CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Summary of the major observations and theories leading from classical theory to quantum theory. Since matter is discontinuous and particulate perhaps energy is discontinuous and particulate. Observation blackbody radiation 7-25 Theory photoelectric effect Energy is quantized; only certain values allowed Einstein: Light has particulate behavior (photons) atomic line spectra Bohr: Dr. Wolf’s CHM 101 Planck: Energy of atoms is quantized; photon emitted when electron changes orbit. Since energy is wavelike, perhaps matter is wavelike Observation Davisson/Germer: electron diffraction by metal crystal Theory 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 Einstein/deBroglie: Mass and energy are wavelength increases equivalent; particles have wavelength and (momentum decreases) photons have momentum. after colliding with electron QUANTUM THEORY Energy same as Matter particulate, massive, wavelike 7-26 Dr. Wolf’s CHM 101 The Heisenberg Uncertainty Principle Heisenberg Uncertainty Principle expresses a limitation on accuracy of simultaneous measurement of observables such as the position and the momentum of a particle. x 7-27 Dr. Wolf’s CHM 101 .mu h 4p Sample Problem 7.4 PROBLEM: PLAN: Applying the Uncertainty Principle An electron moving near an atomic nucleus has a speed 6 x 106 ± 1% m/s. What is the uncertainty in its position ( x)? The uncertainty ( u) is given as ±1% (0.01) of 6 x 106 m/s. Once we calculate this, plug it into the uncertainty equation. SOLUTION: u = (0.01) (6 x 106 m/s) = 6 x 104 m/s x . m u h 4p 6.626 x 10-34 kg-m2/s x 4p (9.11 x 10-31 kg) (6 x104 m/s) 7-28 Dr. Wolf’s CHM 101 10-9m 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. n the principal quantum number - a positive integer l the angular momentum quantum number - an integer from 0 to n-1 ml the magnetic moment quantum number - an integer from -l to +l ms the spin quantum number, + 1/2 or - 1/2 7-29 Dr. Wolf’s CHM 101 Electron probability in the ground-state H atom n=1 l=0 m=0 7-30 Dr. Wolf’s CHM 101 “Orbital” showing 90% of electron probability Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Principal, n Positive integer (size, energy) (1, 2, 3, ...) Angular momentum, l 0 to n-1 (shape) sublevel names l = 0, called “s” l = 1, “p” l = 2, “d” l = 3, “f” Quantum Numbers 1 2 0 0 0 0 3 1 0 -1 0 +1 -l,…,0,…,+l -2 7-31 Dr. Wolf’s CHM 101 2 0 -1 0 +1 Magnetic, ml (orientation) 1 -1 0 +1 +2 Sample Problem 7.5 Determining Quantum Numbers for an Energy Level 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? 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 ml values and therefore 9 orbitals with n = 3. 7-32 Dr. Wolf’s CHM 101 Sample Problem 7.6 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 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 (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 7-33 Dr. Wolf’s CHM 101 sublevel name possible ml values # of orbitals s orbitals 1s 7-34 Dr. Wolf’s CHM 101 2s 3s p orbitals - three of them Combination The 2p orbitals n = 2, l = 1 7-35 Dr. Wolf’s CHM 101 d orbitals - five of them The 3d orbitals n = 3 l = 2 7-36 Dr. Wolf’s CHM 101 d orbitals - five of them Combination 7-37 Dr. Wolf’s CHM 101 f orbitals - seven of them One of the seven possible 4f orbitals 7-38 Dr. Wolf’s CHM 101 End of Chapter 7 7-39 Dr. Wolf’s CHM 101