Question

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

Answer 1

Number of atoms per unit cell in BCC lattice = 2

If half of the atoms are removed, number of remaining atoms per unit cell = 1


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Answer 2

Answer:

The percentage of void is 66%.

Explanation:

In a BCC (Body Centered Cubic) unit cell the effective number of atoms = 2.

This is because, the atoms presents at the vertices share $1 / 4^{\mathrm{th}}$ part with each adjacent lattice and one atom is present at the center of the cube. Thus, number of effective atoms present = (1/4).4 + 1 = 2

The formula for calculating packing efficiency =  $\bold{\frac{\text {Volume} \text { of one atom }}{\text {Volume of unit cell}}}$

The percentage occupied in a BCC lattice = 68%

For 1 atom = 34%

Therefore, percentage void  = 100 – 34 = 66%

Explanation: