00:01
Hi, in this question based on fcc unit less lattice of diamond we have to answer the following.
00:06
Firstly, we need to find out how many atoms are present per unit cells of diamond.
00:12
Number of atoms contributed by center atoms to the unit cell will be equal to 1 by 8 multiplied by 8.
00:33
It comes out one atom.
00:35
Then number of atoms contributed by face center atoms that is 1 by 2 multiplied by 6.
00:53
It comes out three atoms.
00:57
The one marked with red are the face centered atoms here.
01:01
Additional carbon atoms located at four sides are four atoms.
01:17
So total atoms become one plus three plus four.
01:26
It comes out eight atoms per unit cell.
01:30
Now in the second part, we have to provide where are the atoms present.
01:36
The one marked with red are atoms bounded to other four.
01:50
Then these are face centered.
01:56
These are placed at corners and the one marked with this symbol are the atoms bounded outside of cube.
02:08
Now we have to derive relationship between lattice parameter and atomic radius.
02:20
Consider triangle abc where we will be using pythagoras theorem.
02:25
Ab square is equal to ac square plus bc square.
02:29
Length of ac is 1 by 4 square and bc is 1 by 4 square.
02:35
From here we will find out the value of ab...