Question

In a ccp type structure if half of the face centred atoms are removed then % void is?

Answer 1

Answer:

The atoms are present at both the corner and faces in a CCP structure.

Hence,

the present number of atoms are :-

$1/8 * 8 + 1/2 * 6 = 4$

On removal of half of the face centered atoms,

the number of atoms become

$1/8 * 8 + 1/2 * 3 = 5/2$

taking a as edge length, Volume of cube = a$a^3$

Volume occupied by atom becomes = $43\pi r ^ 3$

For FCC lattice,

4r = root(2) * aPacking efficiency

= (number of present atoms*volume of one atoms) / total volume of cube * 100

= $(5/2*43\pi r^3)/ a^3 * 100$

= 46.4%

Hence, the % of void becomes

100 - 46.4 = 53.6 %

Explanation:

Answer 2

Explanation:

hope it will help.................................