# suppose A is the atom present at the corner of f.c.c. cube unit cell
As atom at corner of unit cell is shared by 8 different unit cell hence only 1/8 th part of each corner atom present in one unit cell
in f. c. c. total 8 atoms are present at corners ND 1 corner consist of 1/8 part of atom But 1 atom is missing at corner hence only 7. R at corner. hence total 7 corners consist of ➡ A= 1 / 8 ✖ 7 = 7 / 8 atoms
In the f. c. c. unit cell there are total six faces. and B atoms are present at center of these faces
atom at center of each face of f. c. c. is shared by two unit cells and contribute 1/2 part to each unit cell . hence total atoms at 6 facec in f. c. c. ➡ B = 1 / 2 ✖ 6 = 3
# therefore A = 7/ 8 ND B = 3 hence A : B = 7/8 : 3 = 7 : 24 hence A7 B24 is the answer
for 8 th question I have already given answer in another your question.
Answers & Comments
# suppose A is the atom present at the corner of f.c.c. cube unit cell
As atom at corner of unit cell is shared by 8 different unit cell
hence only 1/8 th part of each corner atom present in one unit cell
in f. c. c. total 8 atoms are present at corners
ND 1 corner consist of 1/8 part of atom
But 1 atom is missing at corner hence only 7. R at corner.
hence total 7 corners consist of
➡ A= 1 / 8 ✖ 7 = 7 / 8 atoms
In the f. c. c. unit cell there are total six faces.
and B atoms are present at center of these faces
atom at center of each face of f. c. c. is shared by two unit cells and contribute 1/2 part to each unit cell .
hence total atoms at 6 facec in f. c. c.
➡ B = 1 / 2 ✖ 6 = 3
# therefore A = 7/ 8 ND B = 3
hence A : B = 7/8 : 3 = 7 : 24
hence A7 B24 is the answer
for 8 th question I have already given answer in another your question.