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## Chapter 7

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**Chapter 7**Quantum Theory and Atomic Structure**Quantum Theory and Atomic Structure**7.1 The Nature of Light 7.2Atomic Spectra 7.3 The Wave-Particle Duality of Matter and Energy 7.4The Quantum-Mechanical Model of the Atom**The Wave Nature of Light**Wave properties are described by two interdependent variables: Frequency: n(nu) = the number of cycles the wave undergoes per second (units of s-1 or hertz (Hz)) (cycles/s) Wavelength: l (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 (10-12 m) or angstroms (Å, 10-10 m) per cycle) (m/cycle) Speed of a wave = cycles/s x m/cycle = m/s c= speed of light in a vacuum = nl =3.00 x 108 m/s**Figure 7.1**Frequency and Wavelength**Amplitude (intensity)**of a Wave (a measure of the strength of the wave’s electric and magnetic fields) Figure 7.2**Regions of the Electromagnetic Spectrum**Figure 7.3 Light waves all travel at the same speed through a vacuum but differ in frequency and, therefore, in wavelength.**PROBLEM:**A dental hygienist uses x-rays (l = 1.00 Å) to take a series of dental radiographs while the patient listens to a radio station (l = 325 cm) and looks out the window at the blue sky (l = 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 108m/s. 1.00 Å 10-10m 1 Å 325 cm 10-2m 1 cm 10-9m 473 nm 1 nm 3.00 x 108m/s 473 x 10-9m Sample Problem 7.1 Interconverting Wavelength and Frequency PLAN: Use c = ln x = 1.00 x 10-10m 3.00 x 108m/s SOLUTION: n = = 3 x 1018s-1 1.00 x 10-10m x = 325 x 10-2m 3.00 x 108m/s n = = 9.23 x 107s-1 325 x 10-2m = 473 x 10-9m x n = = 6.34 x 1014s-1**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**Different behaviors of waves and particles**Figure 7.4**The diffraction pattern caused by light passing through two**adjacent slits Figure 7.5**Blackbody Radiation**Changes in the intensity and wavelength of emitted light when an object is heated at different temperatures Figure 7.6 Planck’s equation E = nhn E = energy of radiation h = Planck’s constant n = frequency n = positive integer (a quantum number) h = 6.626 x 10-34 J.s**The Notion of Quantized Energy**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 hn. An atom changes its energy state by absorbing or emitting one or more quanta of energy. DEatom = Eemitted(or absorbed) radiation= Dnhn DE = hn (Dn = 1)energy change between adjacent energy states**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: Ephoton = hn = DEatom Figure 7.7**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? 10-2m cm Calculating the Energy of Radiation from its Wavelength Sample Problem 7.2 PLAN: After converting cm to m, we use the energy equation, E = hn combined with n = c/l to find the energy. SOLUTION: E = hc/l (6.626 x 10-34J.s) x (3.00 x 108m/s) E = = 1.66 x 10-23J 1.20 cm x**Atomic Spectra**Line spectra of several elements Figure 7.8**1**1 1 l n12 n22 = R - Rydberg equation: R is the Rydberg constant = 1.096776 x 107 m-1 Three series of spectral lines of atomic hydrogen for the visible series, n1 = 2 and n2 = 3, 4, 5, ... Figure 7.9**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**Quantum staircase**Figure 7.10**Limitations of the Bohr Model**Can only predict spectral lines for the hydrogen atom (a one electron model) For systems having >1 electron, there are additional nucleus-electron attractions and electron-electron repulsions Electrons do not travel in fixed orbits**The Bohr explanation of the three series of spectral lines**for atomic hydrogen Figure 7.11**Figure B7.1**Flame tests strontium 38Sr copper 29Cu Emission and absorption spectra of sodium atoms Figure B7.2**Fireworks emissions**from compounds containing specific elements**Samplein compartment absorbs characteristic amount of each**incoming wavelength. Computer converts signal into displayed data. Lenses/slits/collimaters narrow and align beam. Monochromator (wavelength selector) disperses incoming radiation into continuum of component wavelengths that are scanned or individually selected. Sourceproduces radiation in region of interest. Must be stable and reproducible. In most cases, the source emits many wavelengths. Detector converts transmitted radiation into amplified electrical signal. Figure B7.3 The main components of a typical spectrophotometer**The Absorption**Spectrum of Chlorophyll a Absorbs red and blue wavelengths but no green and yellow wavelengths; leaf appears green.**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.**Wave motion in restricted systems**Figure 7.13**h**/mu l = The de Broglie Wavelength Table 7.1 The de Broglie Wavelengths of Several Objects Substance Mass (g) Speed (m/s) l (m) slow electron 9 x 10-28 1.0 7 x 10-4 fast electron 9 x 10-28 5.9 x 106 1 x 10-10 alpha particle 6.6 x 10-24 1.5 x 107 7 x 10-15 one-gram mass 1.0 0.01 7 x 10-29 baseball 142 25.0 2 x 10-34 Earth 6.0 x 1027 3.0 x 104 4 x 10-63**PROBLEM:**Find the deBroglie wavelength of an electron with a speed of 1.00 x 106m/s (electron mass = 9.11 x 10-31kg; h = 6.626 x 10-34kg.m2/s). Calculating the de Broglie Wavelength of an Electron Sample Problem 7.3 PLAN: Knowing the mass and the speed of the electron allows use of the equation, l = h/mu, to find the wavelength. SOLUTION: 6.626 x 10-34kg.m2/s = 7.27 x 10-10m l = 9.11 x 10-31kg x 1.00 x 106m/s**Comparing the diffraction patterns of x-rays and electrons**Figure 7.14 Electrons - particles with mass and charge - create diffraction patterns in a manner similar to electromagnetic waves!**Since matter is discontinuous and particulate, perhaps**energy is discontinuous and particulate. blackbody radiation Planck: Energy is quantized; only certain values are allowed photoelectric effect Einstein: Light has particulate behavior (photons) atomic line spectra Bohr: Energy of atoms is quantized; photon emitted when electron changes orbit. Figure 7.15 CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Summary of the major observations and theories leading from classical theory to quantum theory Observation Theory**Figure 7.15 (continued)**Since energy is wavelike, perhaps matter is wavelike. Davisson/Germer: electron diffraction by metal crystal 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 wavelength increases (momentum decreases) after colliding with electron Einstein/deBroglie: Mass and energy are equivalent; particles have wavelength and photons have momentum. QUANTUM THEORY Energy same as Matter: particulate, massive, wavelike Observation Theory**h**4p The Heisenberg Uncertainty Principle It is impossible to know simultaneously the exact position and momentum of a particle DxxmDu≥ Dx = the uncertainty in position Du = the uncertainty in speed A smaller Dx dictates a larger Du, and vice versa. Implication: cannot assign fixed paths for electrons; can know the probability of finding an electron in a given region of space**PROBLEM:**An electron moving near an atomic nucleus has a speed 6 x 106 ± 1% m/s. What is the uncertainty in its position (Dx)? h D xx mD u ≥ 4p Sample Problem 7.4 Applying the Uncertainty Principle PLAN: The uncertainty in the speed (Du) is given as ±1% (0.01) of 6 x 106m/s. Once we calculate this value, the uncertainty equation is used to calculate Dx. SOLUTION: Du = (0.01)(6 x 106m/s) = 6 x 104m/s (the uncertainty in speed) 6.626 x 10-34kg.m2/s ≥ 1 x 10-9m Dx ≥ 4p (9.11 x 10-31kg)(6 x 104m/s)**d2Y**d2Y d2Y wave function mass of electron potential energy at x,y,z dx2 dy2 dz2 8p2mQ + + + (E-V(x,y,z)Y(x,y,z) = 0 h2 how y changes in space total quantized energy of the atomic system The Schrödinger Equation HY = EY Each solution to this equation is associated with a given wave function, also called an atomic orbital**Electron probability in the ground-state hydrogen atom**Figure 7.16**Quantum Numbers and Atomic Orbitals**An atomic orbital is specified by three quantum numbers. nthe principal quantum number - a positive integer (energy level) lthe angular momentum quantum number - an integer from 0 to (n-1) mlthe magnetic moment quantum number - an integer from -l to +l**0**1 2 0 -1 0 +1 -1 0 +1 -2 -1 0 +1 +2 Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum Numbers Principal, n (size, energy) Positive integer (1, 2, 3, ...) 1 2 3 Angular momentum, l (shape) 0 to n-1 0 0 1 0 0 Magnetic, ml (orientation) -l,…,0,…,+l**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? Anthony S. Serianni: Sample Problem 7.5 Determining Quantum Numbers for an Energy Level PLAN: Follow the rules for allowable quantum numbers. 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 (s sublevel) For l = 1 ml = -1, 0, or +1 (p sublevel) For l = 2 ml = -2, -1, 0, +1, or +2 (d sublevel) There are 9 mlvalues and therefore 9 orbitals with n = 3**PROBLEM:**Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: Sample Problem 7.6 Determining Sublevel Names and Orbital 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, find ml and the number of orbitals. SOLUTION: n l sublevel name possible ml values no. orbitals (a) 2 3d -2, -1, 0, 1, 2 3 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**S orbitals**1s 3s 2s spherical nodes Figure 7.17**2p orbitals**Figure 7.18 nodal planes**3d orbitals**Figure 7.19 perpendicular nodal planes**One of the seven possible 4f orbitals**Figure 7.20**The energy levels**in the hydrogen atom The energy level depends only on the n value of the orbital Figure 7.21