Question

In a FCC unit cell, if half of the atoms per unit cell are removed, then percentage void is

Answer 1

Answer:

in a FCC unit cell if half of the atoms per unit cell are removed then percentage void is

Explanation:

The percentage of void is

=66%

Answer 2

In a FCC unit cell, if half of the atoms per unit cell are removed, then percentage void is 66%

Explanation

• In face centred cubic (fcc) unit cell, the atoms touchs  each other along the face diagonal of the cube
• In fcc  corners=8 atoms,its contribution is 1/8 ,total atom will be 1
• In each faces 6 atoms are present, its contribution is 1/2,total atom will be 6×1/2 =3
• Total atoms = 1+3=4
• Hence 4r=√2a There are four atoms per unit cell√2a
• Hence fraction of volume occupied will be = $4\frac{\frac{4}{3} \pi r^{2}}\frac{4r3}/\sqrt{2}$
• The percentage occupied in a fcc lattice = 68%
• For 1 atom = 34%
• Therefore, percentage void  = 100 – 34 = 66%

#SPJ3