Question

(a) Derive linear density expressions for FCC (100) and [111] directions in terms of the atomic radius Rand (b) compute linear density values for these two directions for silver. (100): atom/R (111) atom/R (b) (100): 1 ! 1/m (111): i 1/m

Answer 1

Sol's Given that por a fcc crystalline Structure of (100) plane there are a atom situated at the two corners of the cell and I atom at the center of the face. → Each comer atom = 1/2 - We have a atoms = latom. → Fre (100) has ito side length equal to arra. R- Radius of the atom. To find the linear density of the vector (100) using. the below Expression, on the = Number of atoms Centered direction vector. Length of direction vector. dans centered on the = 1 arom. No. of atoms centered on the direction vector length o direction Vector = leve,

shows the different vectors in the The figure chystalline Cell :- LAN (llo) Eta por a fcc crystalline structure of Culd rector. it passes thought the center of the cell acting as a diagonal . » It has two half of atoms in its path hence. Pots Contribution is 1 alom To find the length of the vector (110) using trigonometry deo) - vare 2. aic Standard dimensions of the cube. Q- akva. c= QRrg. in fig.

dello = varte? = VQR 12); Carray? | deno) = UR. ] To find the length of the vector (III) using Pigonometry and the figure l. duel) - do +6² b: 2012. duo = UR. du - Vlur? +282? 1 duo = 2Rv6 The linear density of the vector (111) using the Expression Dalo - - direction vector hength of direction rector.de (Due = latom ZRVG. centered on the Los = No. of atoms .

linear linear density of the vector loo- latom 28.2. density of the rector 111 - Tatom 2016 (b) : To find the density of the rector (100) To find the using the sliver linear relation. 10:00 - larom 289 R= 0.1445 2X0.1445X12. 14100 - 2.46670m? 2011- latom 2RV6 R= 0.1405 2 xo.luus x vt Tom = 1.412 60ml