Question

CALCULATE THE NUMBER OF FRENKEL DEFECTS PRESENT IN A CRYSTAL OF AgCl at 300 K, GIVEN THAT THE MATERIAL HAS A CUBE UNIT CELL OF EDGE 0.555 nm THAT CONTAINS 4 A g ATOMS. ASSUME THAT THE INTERSTITIAL ATOMS OCCUMY ANY OF 8 TETRAHEDRAL SITES IN THE UNIT CELL. Hf = 2.69 x 10^-19 L

Answer 1

number of Frankel defects = (square root of ( N * Ni)) * e^(-dH/2RT)

N = number of atoms present in unit cell

Ni = number of interstial positions

dH = enthalpy of formation for frankel defect

R = gas constant

T = absolute temperature

Number of Frankel defects = (square root of ( N * Ni)) * e^(-dH/2RT)

                                                 = (square root of (4 * 8)) * e^((2.69*10^-19)/(2*8.314*300))

                                                 = 1