(A) For a body-centered cubic $(BCC)$ unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $\sqrt{3} a = 4

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(A) For a body-centered cubic $(BCC)$ unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by $\sqrt{3} a = 4r$. The diameter of the atom $(d)$ is equal to $2r$. From the relation $\sqrt{3} a = 4r$,we can write $2r = \frac{\sqrt{3}}{2} a$. Given $a = 4 \times 10^{-10} \ m$,the diameter $d = 2r = \frac{\sqrt{3}}{2} \times 4 \times 10^{-10} \ m$. $d = 1.732 \times 2 \times 10^{-10} \ m = 3.464 \times 10^{-10} \ m$.

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

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In a body-centered cubic $(BCC)$ unit cell of a metal,the edge length is $4 \times 10^{-10} \ m$. The diameter of the atom will be:

A
$3.46 \times 10^{-10} \ m$
B
$2 \times 10^{-10} \ m$
C
$1 \times 10^{-10} \ m$
D
$1.732 \times 10^{-10} \ m$

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Mathematical analysis of cubic system and Bragg’s equation

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In a body-centered cubic $(BCC)$ unit cell of a metal,the edge length is $4 \times 10^{-10} \ m$. The diameter of the atom will be:

Similar Questions

Sodium metal crystallizes in a body-centred cubic $(bcc)$ lattice with a unit cell edge of $4.29 \ \mathring{A}$. The radius of the sodium atom is approximately ............. $\mathring{A}$.

Calculate the number of atoms in $1 \ g$ of a metal if it forms an $fcc$ crystal structure,given that $\varrho \times a^3 = 1.728 \times 10^{-22} \ g$.

An element with molar mass $2.7 \times 10^{-2} \,kg \,mol^{-1}$ forms a cubic unit cell with edge length $405 \,pm$. If its density is $2.7 \times 10^{3} \,kg \,m^{-3}$, what is the nature of the cubic unit cell?

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