(B) For $fcc$ structure,the number of atoms per unit cell $Z = 4$. Density $d = \frac{Z \cdot M}{a^3 \cdot N_A}$. Rearranging for edge length $a$: $a^

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(B) For $fcc$ structure,the number of atoms per unit cell $Z = 4$. Density $d = \frac{Z \cdot M}{a^3 \cdot N_A}$. Rearranging for edge length $a$: $a^3 = \frac{Z \cdot M}{d \cdot N_A}$. Substituting the values: $a^3 = \frac{4 \times 98.99}{3.4 \times 6.022 \times 10^{23}}$. $a^3 = 193.389 \times 10^{-24} \text{ cm}^3$. $a = \sqrt[3]{193.389} \times 10^{-8} \text{ cm} = 5.783 \times 10^{-8} \text{ cm}$. Since $1 \text{ cm} = 10^8 \mathring A$,$a = 5.783 \mathring A$.

Class 12 Chemistry Solid State Mathematical analysis of cubic system and Bragg’s equation

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$A$ compound has $fcc$ structure. If the density of the unit cell is $3.4 \text{ g cm}^{-3}$,what is the edge length of the unit cell? (Molar mass $= 98.99 \text{ g mol}^{-1}$)

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A
$7.783 \mathring A$
B
$5.783 \mathring A$
C
$8.780 \mathring A$
D
$6.083 \mathring A$

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Mathematical analysis of cubic system and Bragg’s equation

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$A$ compound has $fcc$ structure. If the density of the unit cell is $3.4 \text{ g cm}^{-3}$,what is the edge length of the unit cell? (Molar mass $= 98.99 \text{ g mol}^{-1}$)

Similar Questions

Niobium crystallises in a body-centred cubic $(bcc)$ structure. If the density is $8.55 \ g \ cm^{-3}$, then the atomic radius of niobium is (atomic mass of niobium $= 93 \ u$) (in $pm$)

An element with $BCC$ structure has an edge length of $500 \ pm$. If its density is $4 \ g \ cm^{-3}$, find the atomic mass of the element.

An element crystallizes in an $fcc$ structure with an edge length of $200 \, pm$. If $200 \, g$ of this element contains $24 \times 10^{23}$ atoms,calculate its density in $g \, cm^{-3}$.

Tobacco seed globin forms face-centered cubic crystals with a unit cell dimension of $12.3 \, nm$ and a density of $1.287 \, g \, cm^{-3}$. Its molecular mass will be:

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