Guest

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.

Grade:12

7 Answers

Arpan Sarkar
22 Points
9 years ago
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
Drake
277 Points
9 years ago
No @Arpan Sarkar the atoms are kept at ALTERNATE FACE CENTRES not at all face centres .
sizzi
27 Points
9 years ago
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
Drake
277 Points
9 years ago
Itna gamand? iit top mar hai kya?!
Drake
277 Points
9 years ago
btw could you explain?
Raghav
11 Points
7 years ago
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
Rajesh Kumar Sahoo
11 Points
5 years ago
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..

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free