A) 4, 2, 1
B) 2, 4, 1
C) 4, 1, 2
D) 1, 4, 2
Correct Answer: D
Solution :
[d]In simple cubic unit cell, 8 atoms are present at its 8 corners. Each atom contributes one eighth per unit cell. So, total contribution \[=8\times \frac{1}{8}=1\] |
In fcc unit cell, 8 atoms are present at its 8 corners. Each atom contributes one eighth to the unit cell. So, total contribution \[=8\times \frac{1}{8}=1\] |
6 atoms are present at its 6 corners. |
Each atom contributes one half to the unit cell. |
Total contribution \[=6\times \frac{1}{2}=3\] |
\[\therefore \] Total number of atoms in fcc unit cell \[=1+3=4.\] |
In body unit cell, 8 atoms are present at its 8 corners. |
Each atom contributes one eighth to the unit cell. |
Total contribution \[=8\times \frac{1}{8}=1\] |
1 atom is present at body centre of bcc unit cell. |
The atom contributes fully to the unit cell. |
Total contribution \[=1\times \frac{1}{1}=1\] |
\[\therefore \] Total number of atoms in bcc unit cell \[=1+1=2\] |
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