Solid state - If we have to place atoms on alternate face centers how many atoms would be present in fcc arrangement.

							we will have 8 atoms at the corners so 8x1/8=1 contributing atoms then for the face centres we have 1/2X6=3 for each alternating face centre 1/2 is contributed.Since the body centred atom is already included in the face centres we take the sum 1+3=4. So 4 atoms in an fcc arrangement 

						

							2 atom on  face center @ drake baccho se mat khel solid state main dum hai toh mujse poonch ke dikha koi bhi question solid state ka samaj main aayi baat
						

							Main poochta hun... in a solid `P` atoms are present at the corners and `Q` atom is present at the body centre position. What is formula of compound.. also calculate the formula if 2 corner atoms are replaced by R atoms
						

							Each cube contains 8 corners and 6faces.. At each corner no of atom =1/8, so total atoms at corner (P)=1/8*8=1No of atom in center of body(Q) =1So Formula of Solid is PQ as ratio is 1:1. If 2 corner is occupied by R, then no of atom= 1/8*2=1/4. And Q atoms reduced to 1/8*6=6/8. Thus formula becomes P1 Q6/8 R1/4 =P4Q6R1. (multiplying by 4 in formula). Thank u..