The total volume of atoms each with 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 $\frac{16}{3} π r^{3}$
As we know,

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

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

Hence, the correct option is $D$

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Similar questions

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

Question 1
If the radius of a sphere is 2r, then its volume will be:

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B) $4 π r^{3}$
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D) $\frac{32}{3} π r^{3}$

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The total volume of atoms each with radius r present in fcc unit cell of metal is:

The correct option is D 163πr3As we know,Volume of atom in a cell =43πr3×n (n=4 for fcc)=43πr3×4=163πr3Hence, the correct option is D

The correct option is D $\frac{16}{3} π r^{3}$
As we know,

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

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

Hence, the correct option is $D$