The volume of atoms radius r present in fcc unit cell of metal is-A- 20-3- r 3B- 24-3- r 3C- 12-3- r 3D- 16-3- r3

The correct option is D 16/3π r^3 Volume of atom in a cell =4/3π r^3 × n (n=4 for fcc) =4/3×π r^3 × 4 =16/3π r^3 ...

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Standard XII

Chemistry

Packing Fraction and Efficiency

The volume of...

Question

The volume of atoms (radius r) present in fcc unit cell of metal is:

$\frac{20}{3} π r^{3}$

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$\frac{24}{3} π r^{3}$

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$\frac{12}{3} π r^{3}$

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$\frac{16}{3} π r^{3}$

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Solution

The correct option is D
$\frac{16}{3} π r^{3}$

Volume of atom in a cell $= \frac{4}{3} π r^{3} \times n$ (n=4 for fcc)

$= \frac{4}{3} \times π r^{3} \times 4 = \frac{16}{3} π r^{3}$

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The volume of atoms (radius r) present in fcc unit cell of metal is:

The correct option is D 163πr3 Volume of atom in a cell =43πr3×n (n=4 for fcc) =43×πr3×4=163πr3

The correct option is D
$\frac{16}{3} π r^{3}$

Volume of atom in a cell $= \frac{4}{3} π r^{3} \times n$ (n=4 for fcc)

$= \frac{4}{3} \times π r^{3} \times 4 = \frac{16}{3} π r^{3}$