00:01
In this question, we have given a fcc structure.
00:11
Okay, so here we have some given data, let us write that, that atomic radius, let us suppose this is r and that is equal to 2 .25 armstrong.
00:22
Okay, after that we have assuming atoms as hard spheres and touching each other.
00:27
So, fcc contain atom at corners and at faces we have, we know that about the fcc structure.
00:32
So here, the a which is side of the cube or edge length or we can also say it as a edge length.
00:47
Okay.
00:47
Now, the bd, okay bd means what, let us suppose bd means the diagonal of the fcc structure, let us draw it simply.
01:11
This is just a fcc diagram, only rough diagram this is, okay.
01:16
And we have to assume the atoms are as a hard spheres and touching each other.
01:21
Okay, we have to assume like that.
01:23
So, this is the length, which is called a, okay.
01:27
And the r is the radius of the atoms, right.
01:37
So, here, the diagonal which is root 2a, a is the edge length or side of the cube.
01:44
So, the diagonal can be this diagonal, okay, this diagonal would be equal to how many r? 4 r's, 1 r, 2 r and r that is equal to 4 r.
02:02
And from here, we can say that 4 r will be equal to root 2a.
02:07
So, from here, we will get the a value as equal to 4 upon root 2 r...