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Optical Properties of Materials Atomic and Molecular Origins of Color – 2.1, 2.6, 2.7 Colors originate from the interaction of light waves with matter. Specifically, with electrons in matter. The human eye detects light in the 350350-750nm wavelength range, or ~3.5~3.5-1.5eV (1240=1eV*λ (1240=1eV*λ) The longer the wavelength the lower the energy. E = hv = hc / λ Optical Properties of Materials Sources of color are: 1. Energy absorption – requires a source of light (i.e. electronic transitions, molecular vibrations) 2. Energy emission – can be seen in the dark (i.e. phosphorescence, fluorescence) Transmission or Reflection – The perceived color of the object will be the complementary color of the light absorbed. Ch. 2, #1. If six paints are mixed in equal portions (red, orange, orange, yellow, green, blue violet) the paint is black. However if six lights of the same color color shine at the same spot the spot is white. Why? 1 Optical Properties of Materials Color centers (F(F-centers) – Example of free electrons in solids or liquids. For materials not usually colored, their band gaps are >3 eV, eV, i.e. insulators *Arises because of a site defect in the lattice For example, occurrence in NaCl gives a yellow color, and in KCl a violet color. Optical Properties of Materials The trapped electron is a classic example of a particle in a box. box. The spacing of the energy levels and the color observed depends on the the host crystal and not the source of electrons. N-doped Examples of colored diamonds. Model: Radiation dislodges a carbon atom from the diamond structure and an electron takes its place. B-doped * When the electron escapes the trap (through heating or solar radiation), the electrons and atoms assume more regular lattice positions and the color is lost. 2 In-Class Problem Set (1) Rubies are red because of the absorption of light by Cr3+ ions. The color of emeralds also has the same origin in absorption of light light by Cr3+ ions. Explain whether the crystal field experienced by Cr3+ is less in emerald or in ruby. (2) How do the host lattices influence the energy level spacings (and color) of the free electron? For example, why do the color centers centers in NaCl appear yellow and the color centers in KCl appear violet? (3) Ch. 2, #13. Most clothing stores have fluorescent lighting. In this light, two items can appear to be color matched, but when viewed in sunlight, they are not well matched. Explain. Colors in Metals and Semiconductors Metals – free electron gas; Delocalized electrons span a near continuum of closelyclosely-spaced energy levels at Ef. Æ Metals can absorb all wavelengths (energy) of light T = 0K P(E) T > 0K P(E) = 1 kT + (E-Ef)/ Ef)/kT e(E- 1 Ef E (eV (eV)) * At 0K, states above Ef are empty (dashed line) * At temperatures >0K, states above Ef are partially filled (red), and states below Ef are partially empty (green). 3 Colors in Metals and Semiconductors Semiconductors – The electrons must cross an energy gap, Eg, in order to reach the conduction band. Æ Can absorb all light with E > band gap ZnS,ZnO ZnS,ZnO CdS HgS GaAs Kurt Nassau “The Causes of Color” Colors in Metals and Semiconductors Doped Semiconductors – Semiconductors that have internal atomic impurities. Æ Impurities can lower the energy to cross the band gap n-type p-type LED – pn junction N-doped B-doped Kurt Nassau “The Causes of Color” 4 Optical Properties of Materials Color From Interactions of Light Waves with Bulk Matter Sec. 4.1, 4.2, 4.5 Refraction – Change in speed of light from one material to another; n = v / v’ v’ = index of refraction n1sinθ sinθ1 = n2sinθ sinθ2 n1 n2 n2 > n1 θ1 θ2 Light bends down Fermat’ Fermat’s Principle of Least Time How could this effect lead to color? Optical Properties of Materials Light refraction as a source of color Why are the colors of light separated upon passing through the prism? Refraction angle depends on: 1. Wavelength (shorter is refracted refracted more) 2. Larger n refracted more (depends on identity of material) 5 Optical Properties of Materials Light refraction as a source of color Total internal reflection (n1>n2): Substance Air Ice Water Diamond n 1.0003 1.305 1.33 2.42 Critical Angle(o)* 88.6 50.02 * sin-1(n2/n1); n2 ~ 1 (air) 48.7 24.4 Where/How could this lead to color? Optical Properties of Materials Example: Total Internal Reflection of a Rain Droplet (See also: Diamonds, M.A. White) Perceived position of colors Below 42o, sky is bright – light reflection Above 42o, sky is dark – no reflection 6 Optical Properties of Materials Diffraction of Light Waves as a Source of Color Materials having stacked layers with distances (d) in the visible visible range of light (nm); nλ = 2dsinθ 2dsinθ Colloidal crystals – Opals nλ = 2dsinθ 2dsinθ 270nm spheres – SiO2, polystyrene 7