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Solid state - If we have to place atoms on alternate face centers how many atoms would be present in fcc arrangement. 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
No @Arpan Sarkar the atoms are kept at ALTERNATE FACE CENTRES not at all face centres .
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
Itna gamand? iit top mar hai kya?!
btw could you explain?
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..
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