Question

Molybdenum disilicide is an intermetallic compound used as a heating element in high temperature furnaces. The compound is known to crystallize into a body-centered tetragonal (BCT) cell (Mo takes the BCT positions) as shown. In this cell, a = b = 0.321 nm and c = 0.785 nm. Answer the following:

a. How many atoms per unit cell are there?

b. Assume the atoms touch along the diagonal directions in the base (the [110] direction for example). Calculate the Atomic Packing Factor.

c. Calculate the theoretical density of this substance.

MO OO мо M. MO

Answer 1

Date : Page No. Molybdesicam divili cide (Mosio) Guren, for BCT celle a= b 30.321 mm C = 0.785 am الح(ع) atoms in a cell Total mo Mai atome (Noug atoms_at. ) + No-qatom in centre 8 Corney x 8 + 1 +8)+ 8 1+1 = 2 Total norf fi in the unit cell T (No. of atoms on corner) at Noon atoms I inside telt- -Х-% +2 1+2 =3 So, total no. a atoms. per unit Eh+3_= 5

(6) Solus +0.324 0,321 am MC 8= 0.785 Mo M T 0-1308 -0.785mm 90 Mo Atomic Packing factor Natoms. T. Vatoms Yunit cell Natom Volume of a tony v 11 5x xxxt a tong 5 % 4 3 x x x (0.130 - 5* 4*_* x (2.237 X10 3 0.1406 3. N (Vatom) =0.0468 nm3 atoms

'Date : Page No. Nimited Cenit cell 0.785 x 0.321_x 0.321 = 0.0808 nm3 Now, Atomic Packiy factor 0.046.8 -O. 08.08 6,5785. APA = 0.5785 Density of tenit tell man in unit Cell Volume of that cau mas of ane atom = @romeonartM) Apagadrot No, Nat he Density of init cell M atoms Venit cell & Na

Date : Page No. -22 mmo 1.5 932871 710 = 15,932 X10-239 msia 4.66 37066 -23 X 10 Density 2x1015.9) 7(3 * 4.66 663)]*10-23 grown 0.09.08 non 3 (318+13.909) K10 23 Jm 010809 nms 566.69x10-23 gm ༡༢ ༡) Danity S.566 x 10-20 g 아 nom?