Total volume of atoms present in a face-centre cubic (FCC) unit cell of a metal is ( r is atomic radius): - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live | ExamTest.live

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Total volume of atoms present in a face-centre cubic (FCC) unit cell of a metal is (r is atomic radius):

A $\frac{20}{3} π r^{3}$
B $\frac{24}{3} π r^{3}$
C $\frac{12}{3} π r^{3}$
D $\frac{16}{3} π r^{3}$

Solution & Step-by-step Explanation

In an FCC unit cell:Number of atoms per unit cell (Z) = 8 \times \frac{1}{8} (corners) + 6 \times \frac{1}{2} (faces) = 1 + 3 = 4 .Volume of one atom (spherical) = \frac{4}{3} π r^{3} .Total volume of atoms = Z \times Volume of one atom $ Total Volume = 4 \times \frac{4}{3} π r^{3} = \frac{16}{3} π r^{3} $

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Total volume of atoms present in a face-centre cubic (FCC) unit cell of a metal is (r is atomic radius):

A

\frac{20}{3} π r^{3}

B

\frac{24}{3} π r^{3}

C

\frac{12}{3} π r^{3}

D

\frac{16}{3} π r^{3}

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