Question

In a bcc unit cell,if half of the atoms per unit cell are removed,then %age void is???​

Answer 1

Answer:

Void %age = 100 - 34 = 66%

Explanation:

No. of atoms per unit cell in case of BCC = 2

If 50% of atoms will be removed, no. of remaining atoms = 1

packing efficiency = no. of atoms x volume of one atom/volume of a unit cell

                                   = 1 x √ 3/ 16 πa^3  / a^3

                                    = 0.34 = 34%

Void %age = 100 - 34 = 66%

Answer 2

Answer:

Total number of the atoms per unit is also called as BCC lattice =2 and when half of the atom are removed and number of remaining atoms per unit cell =1. Here the BCC unit cell is number of atom per cell volume of 1 atoms / volume of unit cell. Here the answer is 34%. Hope the student can submit the value and start solve to get answer for the given value.