title: Electronic Handbook of Optical Constants of Solids (v1-5)
author: Edward D. Palik
description: "The amount of detail that it contains is truly astonishing...There is a fundamental reason why the HOC [Handbook of Optical Constants of Solids] is much more useful than an abbreviated compilation." -- <i>Philip Baumeister, Coherent Auburn Group</i><br /><br />"The beauty of HOC is in the breadth of the spectral range covered." -- <i>David M. Roessler, General Motors Research Laboratories</i><br /><br />"This set of five volumes...is a unique resource for any science and technology library. It provides materials for researchers and optical device designers with reference facts in a context not available anywhere else...The community of materials researchers, spectroscopists, and optical device designers has reason to welcome the five-volume reference tool...These five books will be an invaluable addition to any researcher's library." -- <i>Ivan P. Kaminow, 1996 IEEE Congressional Fellow On The Staff Of The House of Science Committee, Member Of The National Academy Of Engineering</i>
zlib_text: Chapter 41     p. 1 
  CHAPTER 41 – Molecules and Solids 
   
  Note: 
  At the atomic scale, it is most convenient to have energies in electron-volts and wavelengths in  
   
   
  nanometers.  A useful expression for the energy of a photon in terms of its wavelength is 
   
   
   
  E = hf = hc/λ = (6.63 × 10–34 J · s)(3.00 × 108 m/s)(109 nm/m)/(1.60 × 10–19 J/eV)λ; 
   
   
   
  E = (1.24 × 103 eV · nm)/λ. 
   
   
   
  A factor that appears in the analysis of electron energies is 
   
   
   
  e2/4pÅ0 = (1.60 × 10–19 C)2/4p(8.85 × 10–12 C2/N · m2) = 2.30 × 10–28 J · m. 
   
  1. 
  With the reference level at infinity, the binding energy of the two ions is 
   
   
  Binding energy  = – U = e2/4pÅ0r 
   
   
   
   
   
   
  = (2.30 × 10–28 J · m)/(0.28 × 10–9 m) = 8.21 × 10–19 J =     5.1 eV. 
   
  2. 
  We convert the units: 
   
   
  1 kcal/mol  = (1 kcal/mol)(4186 J/kcal)/(6.02 × 1023 molecules/mol)(1.60 × 10–19 J/eV)  
   
   
   
   
   
  =      0.0435 eV/molecule. 
   
  For KCl we have 
   
   
  (4.43 eV/molecule)[(1 kcal/mol)/(0.0435 eV/molecule)] =      102 kcal/mol.