Presentation on theme: "7-1 Dr. Wolf’s CHM 101 Chapter 7 Quantum Theory and Atomic Structure."— Presentation transcript:
7-1 Dr. Wolf’s CHM 101 Chapter 7 Quantum Theory and Atomic Structure
7-2 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
7-3 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 x cm/sec in vacuum
7-4 Dr. Wolf’s CHM 101 Amplitude (Intensity) of a Wave
7-5 Dr. Wolf’s CHM 101 Regions of the Electromagnetic Spectrum c = = c / 1000 kHz100 MHz 3 m300 m
7-6 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 = m 1cm = m 1nm = m = c/ Use c = 1.00A 325cm 473nm m 1A m 1cm m 1nm = 1.00x m = 325x10 -2 m = 473x10 -9 m == 3x10 8 m/s 1.00x 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
7-7 Dr. Wolf’s CHM 101 Different behaviors of waves and particles.
7-8 Dr. Wolf’s CHM 101 The diffraction pattern caused by light passing through two adjacent slits.
7-9 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.
7-10 Dr. Wolf’s CHM 101 Blackbody Radiation E = h Electromagnetic Radiation E = hc/ Planck’s constant h = x J-s
7-11 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.
7-12 Dr. Wolf’s CHM 101 Demonstration of the photoelectric effect
7-13 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
7-14 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 = 6.626X J-s3x10 8 m/s 1.20cm m cm x = 1.66x J
7-15 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.
7-16 Dr. Wolf’s CHM 101 The line spectra of several elements
7-17 Dr. Wolf’s CHM 101 = RRydberg equation n22n22 1 n12n12 R is the Rydberg constant = 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?
7-18 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. By assuming the electron traveled in circular orbits, the energy level for each orbits was E = x J / n 2 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= h = hc/ = x J ( 1 / n 2 final - 1 / n 2 initial )
7-19 Dr. Wolf’s CHM 101 Quantum staircase
7-20 Dr. Wolf’s CHM 101 The Bohr explanation of the three series of spectral lines.
7-21 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?
7-22 Dr. Wolf’s CHM 101 Wave motion in restricted systems
7-23 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 9x x x x x x10 4 7x x x x x x E = hc/ h/m u so = hc/ E but E = m u c de Broglie Wavelength - giving particles wave-like properties
7-24 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 = 9.11x kg; h = 6.626x 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. = 6.626x kg-m 2 /s 9.11x kgx1.00x10 6 m/s = 7.27x m
7-25 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.
7-26 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
7-27 Dr. Wolf’s CHM 101 The Heisenberg Uncertainty Principle x m u h 44 Heisenberg Uncertainty Principle expresses a limitation on accuracy of simultaneous measurement of observables such as the position and the momentum of a particle..
7-28 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 44 x 4 (9.11 x kg) (6 x10 4 m/s) x kg-m 2 /s m.
7-29 Dr. Wolf’s CHM 101 The Schrödinger Equation - Quantum Numbers and Atomic Orbitals 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 m l the magnetic moment quantum number - an integer from - l to + l 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.
7-30 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
7-31 Dr. Wolf’s CHM 101 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 sublevel names l = 0, called “s” l = 1, “p” l = 2, “d” l = 3, “f”
7-32 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.
7-33 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, , 0, 1 -3, -2, -1, 0, 1, 2, 3 n
7-34 Dr. Wolf’s CHM 101 s orbitals 1s2s3s
7-35 Dr. Wolf’s CHM 101 The 2p orbitals n = 2, l = 1 p orbitals - three of them Combination
7-36 Dr. Wolf’s CHM 101 The 3d orbitals n = 3 l = 2 d orbitals - five of them
7-37 Dr. Wolf’s CHM 101 d orbitals - five of them Combination
7-38 Dr. Wolf’s CHM 101 One of the seven possible 4f orbitals f orbitals - seven of them