Question

Derive the planar density expressions for BCC (111) and (110) planes in terms of the atomic radius R. Compute the planar density values for these two planes for chromium (Cr).

Answer 1

Planas dereity for Bcc C): Salution Belou a heithin nthich is shown a H9ure of Bcc unit ceal prane A I k fig.b The centres of theee A, B and c Lie on Cornes atoms this plane. Also ) planes does net pass through centse t ateom D, whieh is docated at unit cell centr the triangte wLith, A, B and c at its thes In Agure Coxners is an EQuilatesal toiangle so area i angu a )C) alao, t iangle edge langih the longth of a fece diageral xis egual to as indicated in g aV2 2-92t92- 20 and we knouw that For BCC 4R therefoxe Aleo from Ag b) 4 Substitute Eqh 4RES - 4RE 2. 50, area of driang Area: y4 x 4RVZ 2 S0, plaras density fos ) Plane for ecc is o.5atam (BCC) - 0:11 R2 16R2

Alomic sadius for chao mium 0.t50 nm A1SO, -2. 6.4055 nm V3 16R2 R PD Cr) 16A0-1302 PrO.6405x10 m planas density fos 8cc 10) Plane feg- A Bcc unit cea is dsauon in fig for (i1o) Plane there is one atom a cach of the fous cube corners thraugh it passes eash of which is shased ith four adjacent unit cell, aslile centre atom Lies ueith in he unit cell. gheus there is equi vadence of 2atoms asociated uith this Bec 1o) plane From fig, Area of reatang antt cetl edge langth wtieh he eength es Know is for Bcc V3 i8 eual to 4R 4R we Aiagonal Now, from fig. 6 d 4RI4R 12 4RE N3 Avea S0, 4R x 4RVZ V3 = 16R2V2 Now planar deneiy far t10 is - no. of atomu entered on o) plane area of plane PD10 2atons -- 3 for cT, RE Del 30 nm -2 15.C9 nm a2/1.569m Xlo PD110 9xD-130 J2