Find the void volume of fcc unit cell in {cm}³ if the volume of fcc unit cell 1.25 × 10^{-22} {~cm}³.

Correct Answer is : 3.25×10−233.25 × 10^{-23}3.25×10−23

Correct Answer is : 3.25×10−233.25 × 10^{-23}3.25×10−23

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MHTCET 16 May 2024 Shift 1 Paper

Question : 73 of 150

Marks: +1, -0

\mathrm{cm}^{3}

1.25 \times 10^{-22} \mathrm{~cm}^{3}

Solution:

Concept:

In a face-centred cubic (fcc) unit cell, atoms occupy only

74\%

of the total volume, so the remaining

26\%

is the void volume.

Explanation:

For an fcc unit cell, the packing efficiency is fixed at

74\%

This means only

74\%

of the unit cell volume is occupied by atoms.

The rest of the space is unoccupied and is called the void volume.

Percentage of void space

= 100\% - 74\% = 26\%

Given, total volume of the fcc unit cell

= 1.25 \times 10^{-22} \, \mathrm{cm}^3

Void volume

= 26\%

of the total unit cell volume

Void volume

= 1.25 \times 10^{-22} \times \frac{26}{100}

Void volume

= \frac{1.25 \times 26}{100} \times 10^{-22}

Void volume

= 0.325 \times 10^{-22} \, \mathrm{cm}^3

Void volume

= 3.25 \times 10^{-23} \, \mathrm{cm}^3

Answer:

The void volume of the fcc unit cell is

3.25 \times 10^{-23} \, \mathrm{cm}^3

Therefore, the correct option is A.

3.25 \times 10^{-23}

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