The shortest distance between 1st and 5th layer of hcp arrangement is
Answers
The atom in H.C.P. structure are arranged in equidistant parallel planes, in each of which every atom has six equidistant neighbours.we created first one layer of atoms and then a second layer of atoms for face-centred cubic structure. Now, for hexagonal close-packed crystal structure, we do not create a third layer. Instead, the third layer is simply the repetitive of the first layer, the fourth layer is the repetitive of second layer and the 5th layer is the repetitive of 3rd layer so on .
If the centres of the first layer is denoted by points marked A, the second layer is denoted by the points marked B, then each points B is at the top of a regular tetrahedron of which three points marked A.
In the H.C.P. structure, let us assume that the radius of each sphere be r, then the length of each side of the tetrahedron will be 2r.
Therefore, the standing height of the tetrahedron is as 2r Sin 600 =3√ r
The distance between the centroid and midpoint of the any side of the basal triangle is 13 . 3 −−√r =r3√
Hence height of the tetrahedron is (3√−−−−√r ) 2 − (r3√ ) 2 = 2 ( 2 r 23) 1/2
If the spacing between first and third layer is represented by c, then
ca = 4 { ( 2 r 23 ) 1/2 } / 2 r =8/3−−−√ = 1.632