(A) For an $FCC$ unit cell,the number of atoms per unit cell $(n)$ is $4$. Density formula: $\rho = \frac{n \times M}{a^3 \times N_A}$ Rearranging for

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(A) For an $FCC$ unit cell,the number of atoms per unit cell $(n)$ is $4$. Density formula: $\rho = \frac{n \times M}{a^3 \times N_A}$ Rearranging for edge length $(a)$: $a^3 = \frac{n \times M}{\rho \times N_A} = \frac{4 \times 60 \ g \ mol^{-1}}{6.23 \ g \ cm^{-3} \times 6.02 \times 10^{23} \ mol^{-1}}$ $a^3 = \frac{240}{37.5046 \times 10^{23}} \ cm^3$ $a^3 \approx 6.4 \times 10^{-23} \ cm^3 = 64 \times 10^{-24} \ cm^3$ Taking the cube root: $a = \sqrt[3]{64 \times 10^{-24}} \ cm = 4.0 \times 10^{-8} \ cm$.

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

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An element (atomic mass $= 60$) has a face-centered cubic $(FCC)$ structure with a density of $6.23 \ g \ cm^{-3}$. The edge length of the unit cell is.... $(N_A = 6.02 \times 10^{23} \ mol^{-1})$

A
$4.0 \times 10^{-8} \ cm$
B
$4.9 \times 10^{-8} \ cm$
C
$5.3 \times 10^{-8} \ cm$
D
$4.4 \times 10^{-8} \ cm$

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Mathematical analysis of cubic system and Bragg’s equation

If a metal crystallises in $bcc$ structure with edge length of unit cell $4.29 \times 10^{-8} \ cm$,the radius of the metal atom is:

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Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

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Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

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Calculate the number of unit cells in $0.5 \text{ g}$ of a metal if the product of the volume and density of the unit cell is $6.65 \times 10^{-23} \text{ g}$.

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An element (atomic mass $= 60$) has a face-centered cubic $(FCC)$ structure with a density of $6.23 \ g \ cm^{-3}$. The edge length of the unit cell is.... $(N_A = 6.02 \times 10^{23} \ mol^{-1})$

Similar Questions

If a metal crystallises in $bcc$ structure with edge length of unit cell $4.29 \times 10^{-8} \ cm$,the radius of the metal atom is:

The number of atoms in $2.4 \ g$ of body-centred cubic $(BCC)$ crystal with edge length $200 \ pm$ is (density = $10 \ g \ cm^{-3}$,$N_A = 6 \times 10^{23} \ atoms \ mol^{-1}$)

Calculate the number of atoms in $0.3 \ g$ of a metal if it forms a $bcc$ structure,given that $[\rho \times a^3 = 3 \times 10^{-22} \ g]$.

Calculate the number of atoms in $5.4 \ g$ of a metal forming an $fcc$ structure,given that the unit cell volume $\left(a^3\right)$ multiplied by density $\left(\varrho\right)$ is $7.2 \times 10^{-22} \ g$.

Calculate the number of unit cells in $0.5 \text{ g}$ of a metal if the product of the volume and density of the unit cell is $6.65 \times 10^{-23} \text{ g}$.

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