F-CENTER ELECTRON QUANTIZATION IN ALKALI HALIDE CRYSTALS Background Rocksalt structureCsCl structureFluorite StructureRbBrTable 2. Ionic Radii Experimental Procedure Irradiation-produced F-centers in Halide Crystals ReferencesD E P A R T M E N T O F M A T E R I A L S S C I E N C E A N D E N G I N E E R I N G M A S S A C H U S E T T S I N S T I T U T E O F T E C H N O L O G Y 3.014 Materials Laboratory Fall 2006 Experiment 3: Modules α1 F-CENTER ELECTRON QUANTIZATION IN ALKALI HALIDE CRYSTALS Background Alkali halides and alkaline earth halides are strongly ionic compounds that solidify as crystals. The name for the alkali elements (Li, Na, K, Rb and Cs) comes from the Arabic word alqili (ashes of the salt-wort plant that was once used as a source of sodium and potassium for making glass from sand). The name for the halogen elements (F, Cl, Br, I—and the rare radioactive element astatine) comes from the Greek word halos (ocean) that recalls their dissolved presence (as ions) in seawater. Most alkali halides (LiF through CsF; see Table 1) crystallize in the face-centered cubic rocksalt structure with monovalent ions (cations, anions) having well-defined ion radii (Table 2). Because of cation/anion radius-ratio packing issues, CsCl, CsBr and CsI instead crystallize in the simple-cubic CsCl structure. The alkaline earth fluorides CaF2, SrF2 and BaF2 crystallize in the cubic fluorite structure. Ionic radii depend somewhat on coordination. In the rocksalt structure, each ion is 6-coordinated by ions of the opposite charge; in the CsCl structure each ion is 4-coordinated by its opposite ion. Cations in the fluorite structure are 8-coordinated by anions; anions are 4-coordinated by cations. A notable feature of these solids is that they are transparent to electromagnetic radiation with wavelengths extending from the far infrared to the far ultraviolet, and they are thus frequently used for optical windows and lenses, particularly in infrared applications where silicate glasses are not transparent. They have similar refractive indices for visible light (nLiF = 1.39, nNaCl = 1.50, nKCl = 1.49, nKBr = 1.56) to those of the silicate glasses (n ≈ 1.5) more traditionally used for lenses in the visible portion of the electromagnetic spectrum. Their transparency derives from a large band-gap (6-8 eV) between the highest filled states of the valence band (associated with the halogen ions) and the lowest (and unfilled) states of the conduction band (constructed from the empty s-orbitals of the alkali ions). -1-Table 1. Crystal Data for Alkali and Alkaline Earth Halides Compound Lattice parameter, a Compound Lattice parameter, a Rocksalt structure CsCl structure LiF 0.4017 nm CsCl 0.4110 nm LiCl 0.513 CsBr 0.4287 LiBr 0.549 CsI 0.4562 LiI 0.600 NaF 0.462 Fluorite Structure NaCl 0.5628 CaF20.5451 NaBr 0.596 SrF20.578 NaI 0.646 BaF20.6184 KF 0.533 KCl 0.628 KBr 0.6578 KI 0.7052 RbF 0.563 RbCl 0.6571 RbBr 0.6868 RbI 0.7325 CsF 0.601 Source: X-ray Powder Diffraction File (International Centre for X-ray Diffraction, 2005) Table 2. Ionic Radii Ion 4-coordinated, nm 6-coordinated, nm 8-coordinated, nm Li+0.073 0.090 0.106 Na+0.113 0.116 0.132 K+0.151 0.152 0.165 Rb+ 0.166 0.175 Cs+ 0.167 0.181 F−0.117 0.119 Cl− 0.167 Br− 0.182 I− 0.206 Ca2+ 0.114 0.126 Sr2+ 0.132 0.140 Ba2+ 0.149 0.156 Source: R.D. Shannon and C.T. Prewitt, Acta Cryst., 1970, B26, 1046 -2-Traditional chemistry would normally regard these highly ionic compounds as having stoichiometric compositions, deriving from the ion valencies, but this is not necessarily the case. Deviations from stoichiometry are a consequence of chemical equilibrium and the possibility of defects occurring in the crystal structure. In an NaCl crystal, for example, a Na+ cation and a Cl− anion could be removed from the interior and placed on the surface, expanding the crystal and leaving behind in the interior a missing Na+ ion (a cation vacancy, VNa) and a missing Cl− ion (an anion vacancy, VCl). There is an energy cost to form such vacancy defects (called the vacancy formation enthalpy hVf, ≈ 1 eV = 1.6×10−19 J per vacancy in NaCl), but the entropy of the crystal is correspondingly increased with the addition of the defects, and the free energy of the system is consequently lowered from this competition between enthalpy and entropy. Vacancy defects are known as Schottky defects, and the (stoichiometric) pair of defects—a cation and an anion vacancy—formed in this example is called a Schottky pair. The equilibrium concentration cS of such defects depends on temperature cS = c0 exp (-hSf/kBT) (1) where the formation enthalpy hSf is for formation of the Schottky pair of defects (the sum of the cation and anion vacancy formation enthalpies, about twice that for either vacancy or about 2 eV in NaCl), and c0 (~150 for vacancies in NaCl) is an entropic factor mostly governed by changes in the vibrational behavior of the ions surrounding the vacancies. The concentration of such equilibrium defects is not large, ~1 ppm at room temperature in NaCl, about 1 ppt at its melting point of 801˚C. If a NaCl crystal is exposed to sodium vapor (partial pressure pNa), Na atoms can dissolve into the crystal to occupy the equilibrium Na vacancies, with the release of an electron as the Na atom converts to a Na+ ion. Their dissolution leaves an excess of Cl vacancies and thus a non-stoichiometric composition Na1+δCl. The overall reaction chemistry can be expressed as NaCl Na + VNa' + VCl• = NaNa× + VCl• + e' (2) The ', • and × superscripts in eqn. (2) represent the electronic charges associated with the various species, with respect to the NaCl crystal. A vacant Na site (VNa') in the crystal is negatively (') charged because the site is surrounded by Cl− ions uncompensated by the single positive charge that would normally be there but is absent; similarly a Cl vacancy (VCl•) is positively (•) charged. A Na+ ion sitting on a Na site (NaNa×) in the crystal is neutral (×) with respect to what the crystal expects at that site. The non-stoichiometry δ (and hence the excess concentration of VCl over VNa) is related to the Na partial pressure pNa by pNa/pNa˚ = (1/cS˚) {δ + (δ2 + 4cS˚)1/2} (3) or pNa/pNa˚ ≈ δ/cS˚ for δ >> cS˚ , -3-where cS˚ is the equilibrium concentration of Schottky