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7-1 Chapter 7 Quantum Theory and Atomic Structure

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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

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7-3 The Wave Nature of Light Wave properties are described by two interdependent variables: Frequency: (nu) = the number of cycles the wave undergoes per second (units of s -1 or hertz (Hz)) (cycles/s) Wavelength: (lambda) = the distance between any point on a wave and a corresponding point on the next wave (the distance the wave travels during one cycle) (units of meters (m), nanometers (10 -9 m), picometers ( m) or angstroms (Å, m) per cycle) (m/cycle) Speed of a wave = cycles/s x m/cycle = m/s c = speed of light in a vacuum = = 3.00 x 10 8 m/s

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7-4 Figure 7.1 Frequency and Wavelength

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7-5 Figure 7.2 Amplitude (intensity) of a Wave (a measure of the strength of the wave’s electric and magnetic fields)

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7-6 Figure 7.3 Regions of the Electromagnetic Spectrum Light waves all travel at the same speed through a vacuum but differ in frequency and, therefore, in wavelength.

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7-7 Sample Problem 7.1 SOLUTION: PLAN: Interconverting Wavelength and Frequency PROBLEM: A dental hygienist uses x-rays ( = 1.00 Å) 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, 3.00 x 10 8 m/s. Use c = 1.00 Å 325 cm 473 nm m 1 Å m 1 cm m 1 nm = 1.00 x m = 325 x m = 473 x m == 3.00 x 10 8 m/s 1.00 x m = 3 x s -1 == == 3.00 x 10 8 m/s 325 x m = 9.23 x 10 7 s x 10 8 m/s 473 x m = 6.34 x s -1 x x x

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7-8 Distinguishing Between a Wave and a Particle Refraction: the change in the speed of a wave when it passes from one transparent medium to another Diffraction: the bending of a wave when it strikes the edge of an object, as when it passes through a slit; an interference pattern develops if the wave passes through two adjacent slits

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7-9 Figure 7.4 Different behaviors of waves and particles

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7-10 Figure 7.5 The diffraction pattern caused by light passing through two adjacent slits

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7-11 Figure 7.6 Blackbody Radiation E = nh Changes in the intensity and wavelength of emitted light when an object is heated at different temperatures Planck’s equation E = energy of radiation h = Planck’s constant = frequency n = positive integer (a quantum number) h = x J. s

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7-12 If an atom can emit only certain quantities of energy, then the atom can have only certain quantities of energy. Thus, the energy of an atom is quantized. Each energy packet is called a quantum and has energy equal to h. An atom changes its energy state by absorbing or emitting one or more quanta of energy. E atom = E emitted (or absorbed) radiation = nh E = h ( n = 1) energy change between adjacent energy states The Notion of Quantized Energy

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7-13 Figure 7.7 Demonstration of the photoelectric effect Wave model could not explain the (a) presence of a threshold frequency, and (b) the absence of a time lag. Led to Einstein’s photon theory of light: E photon = h = E atom

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7-14 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 1.20 cm. What is the energy of one photon of this microwave radiation? After converting cm to m, we use the energy equation, E = h combined with = c/ to find the energy. E = hc/ E = (6.626 x J. s)(3.00 x 10 8 m/s) 1.20 cm m cm x = 1.66 x J x

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7-15 Figure 7.8 Line spectra of several elements Atomic Spectra

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7-16 =R Rydberg equation: n22n22 1 n12n12 R is the Rydberg constant = x 10 7 m -1 Figure 7.9 Three series of spectral lines of atomic hydrogen for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...

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7-17 The Bohr Model of the Hydrogen Atom Postulates: 1. The H atom has only certain allowable energy levels 2. The atom does not radiate energy while in one of its stationary states 3. The atom changes to another stationary state (i.e., the electron moves to another orbit) only by absorbing or emitting a photon whose energy equals the difference in energy between the two states The quantum number is associated with the radius of an electron orbit; the lower the n value, the smaller the radius of the orbit and the lower the energy level. ground state and excited state

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7-18 Figure 7.10 Quantum staircase

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7-19 Limitations of the Bohr Model Can only predict spectral lines for the hydrogen atom (a one electron model) Electrons do not travel in fixed orbits For systems having >1 electron, there are additional nucleus-electron attractions and electron-electron repulsions

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7-20 Figure 7.11 The Bohr explanation of the three series of spectral lines for atomic hydrogen

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7-21 Figure B7.2 Figure B7.1 Emission and absorption spectra of sodium atoms Flame tests strontium 38 Sr copper 29 Cu

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7-22 Fireworks emissions from compounds containing specific elements

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7-23 Figure B7.3 The main components of a typical spectrophotometer Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected. Sample in compartment absorbs characteristic amount of each incoming wavelength. Computer converts signal into displayed data. Source produces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths. Lenses/slits/collimaters narrow and align beam. Detector converts transmitted radiation into amplified electrical signal.

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7-24 The Absorption Spectrum of Chlorophyll a Absorbs red and blue wavelengths but no green and yellow wavelengths; leaf appears green.

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7-25 The Wave-Particle Duality of Matter and Energy De Broglie: If energy is particle-like, perhaps matter is wavelike! Electrons have wavelike motion and are restricted to orbits of fixed radius; explains why they will have only certain possible frequencies and energies.

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7-26 Figure 7.13 Wave motion in restricted systems

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7-27 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 9 x x x x x x x x x x x x h /mu/mu The de Broglie Wavelength

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7-28 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 1.00 x 10 6 m/s (electron mass = 9.11 x kg; h = x kg. m 2 /s). Knowing the mass and the speed of the electron allows use of the equation, = h/m u, to find the wavelength. = x kg. m 2 /s 9.11 x kgx1.00 x 10 6 m/s = 7.27 x m

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7-29 Figure 7.14 Comparing the diffraction patterns of x-rays and electrons Electrons - particles with mass and charge - create diffraction patterns in a manner similar to electromagnetic waves!

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7-30 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 are 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

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7-31 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

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7-32 The Heisenberg Uncertainty Principle x x m u ≥ h 44 It is impossible to know simultaneously the exact position and momentum of a particle x = the uncertainty in position u = the uncertainty in speed A smaller x dictates a larger u, and vice versa. Implication: cannot assign fixed paths for electrons; can know the probability of finding an electron in a given region of space

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7-33 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 in the speed ( u ) is given as ±1% (0.01) of 6 x 10 6 m/s. Once we calculate this value, the uncertainty equation is used to calculate x. u = (0.01)(6 x 10 6 m/s) = 6 x 10 4 m/s (the uncertainty in speed) x x m u ≥ h 44 x ≥ 4 (9.11 x kg)(6 x 10 4 m/s) x kg. m 2 /s ≥ 1 x m

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7-34 The Schrödinger Equation H = E d2d2 dy 2 d2d2 dx 2 d2d2 dz 2m82m 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 Each solution to this equation is associated with a given wave function, also called an atomic orbital

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7-35 Figure 7.16 Electron probability in the ground-state hydrogen atom

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7-36 Quantum Numbers and Atomic Orbitals An atomic orbital is specified by three quantum numbers. n the principal quantum number - a positive integer (energy level) 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

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7-37 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

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7-38 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. 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 (s sublevel) For l = 1 m l = -1, 0, or +1 (p sublevel) For l = 2 m l = -2, -1, 0, +1, or +2 (d sublevel) There are 9 m l values and therefore 9 orbitals with n = 3 Anthony S. Serianni:

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7-39 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, find m l and the number of orbitals. n l sublevel namepossible m l valuesno. orbitals (a) (b) (c) (d) d3d 2s2s 5p5p 4f4f -2, -1, 0, 1, , 0, 1 -3, -2, -1, 0, 1, 2,

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7-40 Figure s 2s 3s S orbitals spherical nodes

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7-41 Figure p orbitals nodal planes

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7-42 Figure d orbitals perpendicular nodal planes

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7-43 Figure 7.19 (continued)

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7-44 Figure 7.20 One of the seven possible 4f orbitals

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7-45 Figure 7.21 The energy levels in the hydrogen atom The energy level depends only on the n value of the orbital

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