The Quantum Mechanical Model of the Atom Chapter 7
The nature of light • Light is electromagnetic radiation: a wave of oscillating electric & magnetic fields.
Wave properties • A wave has wavelength, frequency, and amplitude
Wave properties • Wavelength = (lambda), in m • Frequency = (nu), in cycles/sec or s-1 • Wavelength & frequency related by wave speed: • Speed of light c = 3.00 x 108 m s-1
Example1 • Calculate the frequency of light with = 589 nm
Example 1 • Calculate the frequency of light with = 589 nm • = c c = 3.00 x 108 m s-1 • Must convert nm to m so units agree (n = 10-9)
Example 2 • Calculate the wavelength in nm of light with = 9.83 x 1014 s-1
Electromagnetic spectrum = all wavelengths of electromagnetic radiation
Photoelectric effect • Light shining on metal surface can cause metal to emit e- (measured as electric current)
Photoelectric effect • Classical theory • more e- emitted if light either brighter (amplitude) or more energetic (shorter ) • e– still emitted in dim light if given enough time for e- to gather enough energy to escape
Photoelectric effect • Observations do not match classical predictions! • A threshold frequency exists for e- emission: no e- emitted below that regardless of brightness • Above threshold , e- emitted even with dim light • No lag time for e- emission in high , dim light
Albert Einstein explainsphotoelectric effect • Light comes in packets or particles called photons • Amount of energy in a photon related to its frequency • h = 6.626 x 10-34 J s
Example 3 • What is the energy of a photon with wavelength 242.4 nm?
Example 4 • A Cl2 molecule has bond energy = 243 kJ/mol. Calculate the minimum photon frequency required to dissociate a Cl2 molecule.
Example 4 • What color is this photon? • A photon with wavelength 492 nm is blue
Example 4 • A Cl2 molecule has bond energy = 243 kJ/mol. Calculate minimum photon frequency to dissociate a Cl2 molecule. What color is this photon? • A photon with wavelength 492 nm is blue
Emission spectra • When atom absorbs energy it may re-emit the energy as light
Emission spectra • White light spectrum is continuous • Atomic emission spectrum is discontinuous • Each substance has a unique line pattern
Hydrogen emission spectrum • Visible lines at • 410 nm (far violet) • 434 nm (violet) • 486 nm (blue-green) • 656 nm (red)
Emission spectra • Classical theories could not explain • Why atomic emission spectra were not continuous • Why electron doesn’t continuously emit energy as it spirals into the nucleus
Bohr model • Niels Bohr’s model to explain atomic spectra • electron = particle in circular orbit around nucleus • Only certain orbits (called stationary states) can exist rn = orbit radius, n = positive integer, a0 = 53 pm • Electron in stationary state has constant energy RH = 2.179 x 10–18 J • Bohr model is quantized
Bohr model • e– can pass only from one allowed orbit to another • When making a transition, a fixed quantum of energy is involved
Bohr model • Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2 • Photon energy is an absolute amount of energy • Electron absorbs photon, ∆Eelectron is + • Electron emits photon, ∆Eelectron is –
Bohr model • Calculate the wavelength of light emitted when the hydrogen electron jumps from n=4 to n=2 • 486 nm corresponds to the blue-green line
Example • What wavelength of light will cause the H electron to jump from n=1 to n=3? To what region of the electromagnetic spectrum does this photon belong?
Example • What wavelength of light will cause the H electron to pass from n=1 to n=3?
Example • What wavelength of light will cause the H electron to pass from n=1 to n=3? • The atom must absorb an ultraviolet photon with = 103 nm
Light has both particle & wave behaviors • Wave nature shown by diffraction
Light has both particle & wave behaviors • Particle nature shown by photoelectric effect
Electrons also have wave properties • Individual electrons exhibit diffraction, like waves • How can e– be both particle & wave?
Complementarity • Without laser, single e– produces diffraction pattern (wave-like) • With laser, single e– makes a flash behind one slit or the other, indicating which slit it went through –– and diffraction pattern is gone (particle-like) • We can never simultaneously see the interference pattern and know which slit the e– goes through
Complementarity • Complementary properties exclude each other • If you know which slit the e– passes through (particle), you lose the diffraction pattern (wave) • If you see interference (wave), you lose information about which slit the e– passes through (particle) • Heisenberg uncertainty principle sets limit on what we can know
Indeterminancy • Classical outcome is predictable from starting conditions • Quantum-mechanical outcome not predictable but we can describe probability region
Electrons & probability • Schrodinger applied wave mechanics to electrons • Equation (wave function, ) describe e– energy • Equation requires 3 integers (quantum numbers) • Plot of 2 gives a probability distribution map of e– location = orbital • Schrodinger wave functions successfully predict energies and spectra for all atoms
Quantum numbers • Principal quantum number, n • Determines size and overall energy of orbital • Positive integer 1, 2, 3 . . . • Corresponds to Bohr energy levels
Quantum numbers • Angular momentum quantum number, l • Determines shape of orbital • Positive integer 0, 1, 2 . . . (n–1) • Corresponds to sublevels
Quantum numbers • Magnetic quantum number, ml • Determines number of orbitals in a sublevel and orientation of each orbital in xyz space • integers –l . . . 0 . . . +l
Shapes of orbitals • s orbital (l = 0, ml = 0) • p orbitals (l = 1, ml = –1, 0, +1)
Shapes of orbitals • d orbitals (l = 2, ml = –2, –1, 0, +1, +2)
What type of orbital is designated by each set of quantum numbers? • n = 5, l = 1, ml = 0 • n = 4, l = 2, ml = –2 • n = 2, l = 0, ml = 0 • Write a set of quantum numbers for each orbital • 4s • 3d • 5p
What type of orbital is designated by each set of quantum numbers? • n = 5, l = 1, ml = 0 5p • n = 4, l = 2, ml = –2 4d • n = 2, l = 0, ml = 0 2s • Write a set of quantum numbers for each orbital • 4s n = 4, l = 0, ml = 0 • 3d n = 3, l = 2, ml = –2, –1, 0, +1, or +2 • 5p n = 5, l = 1, ml = –1, 0, or +1
Electron configurations • Aufbau principle: e– takes lowest available energy • Hund’s rule: if there are 2 or more orbitals of equal energy (degenerate orbitals), e– will occupy all orbitals singly before pairing • Pauli principle: • Adds a 4th quantum number, ms (spin) • No two e– in an atom can have the same set of 4 quantum numbers ⇒ 2 e– per orbital