InterviewSolution
Saved Bookmarks
| 1. |
If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in the cube will be:A. `(2)/(sqrt(3))a`B. `(4)/(sqrt(3))r`C. `(sqrt(3))/(4)a`D. `(sqrt(3))/(2)a` |
|
Answer» Correct Answer - 4 The distance between the body centered atom and one corner atom in the cube is actually the nearest neighbour distance the distance between centres of touching spheres. In body centered cubic unit cell ,the atom at the centre is in touch with the other two atoms diagonally arranged at the corners,if a is the lenght of the side of a cube , the length of a body diagonal is equal to `sqrt(3)a` .it is also equal to `4r`,where `r` is the radius of the sphere (atom) as all the three sphere (atoms) along the diagonal touch each other Therefore `sqrt(3)a=4r` `r=(sqrt(3))/(4)a` The nearest neighbour distance `(d)` is `2r`. Therefore `d=2r=2((sqrt(3))/(4)a)=sqrt(3)/(2)a` Note that nearest neighbouring distance `d` is half of the body diagonal. |
|