If the number of missing anions is more than cation. So, to maintain electrical neutrality, some electrons are trapped at anionic vacancies. Because of these extra electrons, the electrical and optical properties of the compound get modified. So these locations are known as F -centres. For example, in NaCl, sodium atoms get deposited on the surface of the crystal when sodium chloride is heated in an atmosphere of sodium vapour. In this process, the chloride ions get diffused with sodium ions and form sodium chloride. In this process, sodium atoms release electrons to form sodium ions. These released electrons get diffused and occupy the anionic sites in the crystal of sodium chloride and creating anionic vacancies that result in the excess of sodium metal. The anionic site occupied by unpaired electrons is called F-centre. They impart yellow colour when visible light falls over the crystal of NaCl, the unpaired electrons present get excited because of the absorption of energy.

Derive Bragg's equation.

In the following figure of beams, the phases of the beams coincide only when the incident angle equals the reflecting angle. The incident beams are parallel to each other until they reach point Z . When they are at point Z , they strike the surface and travel upwards. At point B the 2nd beam scatters. AB + BC is the distance travelled by the 2nd beam. The extra distance is termed as the integral multiple of the wavelength. nλ = AB + BC We know that AB + BC nλ = 2 AB ------------ (1) Further, D is the hypotenuse of the right ΔABZ . Also AB is the opposite side of θ angle. Hence, AB = d sinθ ------------(2) Substitution of (1) in (2) we get: nλ = 2 d sinθ The equation derived is called Bragg's equation.

Silver crystallizes in FCC lattice. If edge length of the cell is 4.07 × 10 - 8 c m and density is 10.5 g c m - 3 . Calculate the atomic mass of silver.

Given, Edge length of the cell ( a ) = 4.07 × 10 - 8 c m Density d = 10.5 g c m - 3 We also know that: the lattice is FCC type, the number of atoms per unit cell , Z = 4 N A = 6.022 × 10 23 m o le - 1 Using the relation d = Z M a 3 N A Therefore, M = da 3 N A Z = 10 .5 gcm - 3 x 4 .077 × 10 - 8 cm 3 × 6 .022 × 10 23 mole - 1 4 = 107.13 g mole - 1 Therefore, the atomic mass of silver = 107 .13 U