Figure 7.3 Regions of the electromagnetic spectrum.
Sample Problem 7.1 SOLUTION: PLAN: Interconverting Wavelength and Frequency wavelength in units given wavelength in m Use c = = 1.00x10 -10 m = 325x10 -2 m = 473x10 -9 m = 3x10 8 m/s 1.00x10 -10 m = = 3x10 8 m/s 325x10 -2 m frequency (s -1 or Hz) = c/ 10 -2 m 1 cm 10 -9 m 1 nm = 3x10 18 s -1 = 9.23x10 7 s -1 3x10 8 m/s 473x10 -9 m = 6.34x10 14 s -1 PROBLEM: A dental hygienist uses x-rays ( = 1.00A) 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, 3.00x10 8 m/s.) o 1 A = 10 -10 m 1 cm = 10 -2 m 1 nm = 10 -9 m o 325 cm 473nm 1.00A o 10 -10 m 1A o
Figure 7.4 Different behaviors of waves and particles.
Figure 7.5 The diffraction pattern caused by light passing through two adjacent slits.
Figure 7.6 Blackbody radiation E = n h E = n h E = h when n = 1 smoldering coal electric heating element light bulb filament
Figure 7.7 Demonstration of the photoelectric effect.
Sample Problem 7.2 SOLUTION: PLAN: Calculating the Energy of Radiation from Its Wavelength 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*s 3x10 8 m/s 1.20cm x 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? 10 -2 m cm = 1.66x10 -23 J
Figure 7.8 The line spectra of several elements.
Table 7.1 The de Broglie Wavelengths of Several Objects Substance Mass (g) Speed (m/s) (m) slow electron fast electron alpha particle one-gram mass baseball Earth 9x10 -28 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 h /m u
Sample Problem 7.3 SOLUTION: PLAN: Calculating the de Broglie Wavelength of an Electron 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 kg x 1.00x10 6 m/s 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). = 7.27x10 -10 m
Bohr’s Model of The Hydrogen Atom <ul><li>Ground state </li></ul><ul><li>Excited state </li></ul><ul><li>When an electron falls back to the lower level, it emits a definite amount of energy in the form of a quantum of light. </li></ul><ul><li>Ex. X-rays </li></ul>
Limitations of the Bohr Model -> Quantum Mechanical Model <ul><li>Unfortunately, the Bohr Model failed for all other elements that had more than one proton and one electron. (The multiple electron-nuclear attractions, electron–electron repulsions, and nuclear repulsions make other atoms much more complicated than hydrogen.) </li></ul><ul><li>In 1920s, a new discipline, quantum mechanics , was developed to describe the motion of submicroscopic particles confined to tiny regions of space. </li></ul>
CLASSICAL THEORY Matter particulate, massive Energy continuous, wavelike Observation Theory Figure 7.15 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. 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
Observation Theory Figure 7.15 continued Since energy is wavelike perhaps matter is wavelike 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 Observation Theory 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
The Heisenberg Uncertainty Principle x * m u ≥ h 4
The Schrödinger Equation H = E d 2 dy 2 d 2 dx 2 d 2 dz 2 + + 8 2 m h 2 (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,z wave function
Quantum Numbers and Atomic Orbitals 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
Quantum Numbers Energies n 1 < 2 < 3 < 4… l s < p < d < f m l Orbitals in the same subshell have equal energies
Azimuthat quantum number, l Subshell designation Maximum number of electrons per subshell 0 s 2 1 p 6 2 d 10 3 f 14
Table 7.2 The Hierarchy of Quantum Numbers for Atomic Orbitals Name, Symbol (Property) Allowed Values Quantum 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 1 0 0 2 0 1 0 3 0 1 2 0 0 -1 +1 -1 0 +1 0 +1 +2 -1 -2
Sample Problem 7.5 SOLUTION: PLAN: Determining Quantum Numbers for an Energy Level 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 . 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? 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.
Sample Problem 7.6 SOLUTION: PLAN: 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 Combine the n value and l designation to name the sublevel. Knowing l , we can find m l and the number of orbitals. n l sublevel name possible m l values # of orbitals (a) (b) (c) (d) 3 2 5 4 2 0 1 3 3d 2s 5p 4f -2, -1, 0, 1, 2 0 -1, 0, 1 -3, -2, -1, 0, 1, 2, 3 5 1 3 7 PROBLEM: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel with the following quantum numbers: