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Wave Nature of Light Rutherford’s model of the atom could not explain chemical behavior Bohr and others described the arrangement of electrons around the.

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Presentation on theme: "Wave Nature of Light Rutherford’s model of the atom could not explain chemical behavior Bohr and others described the arrangement of electrons around the."— Presentation transcript:

1 Wave Nature of Light Rutherford’s model of the atom could not explain chemical behavior Bohr and others described the arrangement of electrons around the nucleus These arrangements could account for differences between the elements Bohr’s model was based on spectroscopic evidence

2 Electromagnetic Radiation
Light can be described as though it is a wave Parts of a wave Amplitude and wavelength Crest Trough

3 Wave Stuff Amplitude: Vertical distance from crest to midline (brightness/intensity) Wavelength: Horizontal distance from crest to crest (meters) (color) Velocity: Rate at which wave travels (3.00x108 m/s in a vacuum) (constant) Frequency: # waves that pass a given point per unit time (waves/sec, cps, hertz, s-1)




7 Electromagnetic Spectrum


9 High frequency Low frequency


11 Wave Math Velocity = wavelength x frequency Velocity = c
Wavelength = l Frequency = n c = ln Since c is constant, l and n are inversely related

12 Example problems Find the frequency of purple light (wavelength 455 nm). Solution: c = ln, or n = c/l Wavelength must be in meters, since c is in m/s: 455 nm = 4.55x10-7 m n = (3.00x108 m/s)/(4.55x10-7 m) = 6.59x1014s-1

13 Example #2 Find the wavelength of red light with a frequency of 4.56x1014s-1. c = ln or l = c/n l = (3.00x108 m/s)/(4.56x1014s-1) = 6.58x10-7m (658 nm)

14 Particle Nature of Light
Light also acts as a particle: proposed by Newton Max Planck and glowing blackbodies: energy is quantized Smallest energy unit available is a quantum Energy of quanta depends on the frequency of the energy: E = hn

15 Particle Nature of Light
h is Planck’s constant: x10-34Js Energy is directly proportional to frequency Energy is inversely proportional to wavelength: E = hn and n = c/l so E = hc/l

16 Light Particle Sample Problem
Find the energy of a microwave photon having a wavelength of 3.42x10-2m. Solution: E = hc/l = (6.626x10-34Js)(3.00x108m/s)/3.42x10-2m = 5.81x10-24J

17 Photoelectric Effect Certain metals will eject electrons when exposed to light Number of electrons ejected depends only on the intensity of light No electrons ejected by light below certain frequency

18 Photoelectric Effect Could not be explained by wave model of light
Einstein explained effect using quantum theory of light He won the Nobel Prize for his work (not for relativity)

19 Photoelectric Effect Simulator

20 Atomic Emission Spectra
When elements are zapped with energy, they give off light Light is first shone through a slit When light is shone through a prism, colors are separated Only some of the colors appear as fine lines against a dark background


22 Emission spectra of common fluorescent bulbs

23 Emission Spectrum Setup

24 Absorption Spectra In absorption spectra, light is shone through cold gas or liquid Light then goes through a slit and prism or grating Resulting spectrum is continuous except for dark lines

25 Absorption Spectrum

26 Absorption Spectrum of Chlorophyll
Liquids tend to have less distinct absorption spectra

27 Gas Absorption Spectra

28 Absorption Spectrum Setup

29 Spectra Types

30 Bohr Model of the Atom Bohr wanted to explain the presence of sharp lines in the hydrogen spectrum He proposed that hydrogen’s electron could only have certain distinct energies These energies were integral multiples of some minimum energy The energy levels correspond to differently sized orbits

31 Bohr’s Atom Spectral lines were due to electrons jumping from one level to another. Incoming energy promotes an electron to a higher energy level When the electron returns to the lower level it releases energy

32 Quantum Numbers Bohr assigned the energy levels numbers
The Principle Quantum Number (n) represents the main energy level n can only have non-zero integral values n = 1, 2, 3, ...

33 Why quantum numbers? Louis de Broglie: Wave-particle duality
As a wave: E = hc/l As a particle: E = mc 2 Combined: hc/l = mc 2 l = hc/mc 2 = h/mc For objects moving slower than light, replace c with v (velocity): l = h/mv

34 Particle wavelength problems
Every particle has a wavelength The larger the particle, the shorter the wavelength Example: Calculate the wavelength of an electron moving at 0.80c (mass = 9.109×10-31 kilograms). Solution: l = h/mv = 6.626x10-34Js/[(9.109×10-31kg)(0.80)(3.00x108m/s)] = 3.0x10-12m (smaller than an atom, bigger than a nucleus)

35 Particle wavelength problems
Find the wavelength of a baseball (145g) thrown toward home plate at 95.0 mph (42.5 m/s) Solution: l = h/mv = 6.626x10-34Js/[(0.145kg)(42.5m/s)] = 1.08x10-34m (much smaller than a nucleus)

36 Back to quantum numbers
Only certain orbits are allowed because they are the only ones in which an integral number of wavelengths can “fit”. “In-between” orbitals would require a fractional number of wavelengths. “I think it is safe to say that no one understands quantum mechanics.” Physicist Richard P. Feynman

37 Heisenberg Uncertainty Principle
It is impossible to know both the position and momentum of an electron simultaneously Electrons are both particles and waves It’s in their nature to be indeterminate Can be thought of as being “smeared out” over a region of space Indeterminacy is related to Planck’s constant

38 More Energy Levels The fine lines in emission spectra are actually made up of several even finer lines Each energy level has sublevels Each sublevel has a shape Each sublevel has one or more orbitals Each orbital holds two electrons How do we sort all this out?

39 Using Quantum Numbers! Four quantum numbers are needed in Schrödinger’s equation to describe the probability function of an electron n = principle quantum number = 1, 2, 3, ... Main energy level – determines size of orbital l = azimuthal quantum number = 0, 1, ... n-1 Sublevel – determines orbital shape

40 Well-used quantum numbers
s: l = 0 (first two columns of PT) p: l = 1 (last six columns of PT) d: l = 2 (middle ten columns of PT) f: l = 3 (bottom two rows of PT) m = magnetic quantum number = - l to +l Specifies orbital – determines orientation s = spin quantum number = ±½ Specifies spin

41 Orbital shapes s orbital: spherical
Every energy level has an s orbital: 1s, 2s, etc. Higher level s orbitals are lobed Nodes are areas of minimum electron density One node is added for each level s sublevel: one orbital, two electrons

42 p orbital p orbitals are dumbbell shaped
p sublevel (l = 1) consists of three orbitals: px py pz (six electrons) Three p orbitals are orthogonal to each other Only present after first main energy level (n>1)

43 d orbital d orbital is cloverleaf-shaped
Five orbitals, ten electrons make up d sublevel Only available when n>2

44 f orbitals Complicated shape Seven orbitals, fourteen electrons
Only available when n>3



47 Allowed quantum number combinations
Pauli exclusion principle: no two electrons can have the same set of four quantum numbers Aufbau principle: electrons fill the lowest energy state available first Lower numbers mean lower energy (n and l) Various m and s states are degenerate (of equal energy)

48 Allowed Quantum Number Combinations
n l m s ½ ½ 1s sublevel – 2e- ½ ½ 2s sublevel – 2e- n l m s ½ ½ ½ ½ ½ ½ 2p sublevel – 6e-

49 Allowed Quantum Number Combinations
3s and 3p are similar to 2s and 2p 3d sublevel: n l m s ½ ½ ½ ½ n l m s ½ ½ ½ ½ ½ ½ 3d sublevel - 10e-

50 Electron configurations
Electron configurations show the location of every electron in the atom Electrons follow three rules: Pauli exclusion principle, Aufbau principle, Hund’s rule Each orbital is represented by a box and a symbol, and each electron by an arrow.

51 Electron configurations

52 More Electron Configurations
Hund’s rule: When putting electrons into degenerate orbitals, do not pair them until necessary.

53 More Electron Configurations

54 More Electron Configurations


56 Add non-standard configurations
And reason for 4s-3d order

57 Orbital filling diagrams

58 Orbital filling diagrams

59 Noble Gas Shorthand Structures
Noble gas symbols can be used to represent the core electron structure Manganese (25 e-): Equivalent to [Ar] 4s2 3d5 1s2 2s2 2p6 3s2 3p6 4s2 3d5

60 Electron Dot Structures
Electron dot structures represent only the valence electrons Valence electrons are the electrons in the outermost energy level (highest value of n) The maximum number of electrons allowed in the valence shell is 8

61 Electron dot structures
Consists of the element’s symbol and dots representing the valence electrons Hydrogen: Helium: Lithium: Beryllium: Nitrogen: Neon: Sodium: Iron: Lead: H He Li Be N Ne Fe Na Pb

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