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

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(D) For a body-centered cubic $(BCC)$ lattice,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $\sqrt{3} \ a = 4 \ r$.Given $a = 4 \ \mathring{A}$,we substitute the value into the equation:$\sqrt{3} \times 4 = 4 \ r$.$r = \sqrt{3} \ \mathring{A} \approx 1.732 \ \mathring{A}$.To express this in the form $..... \times 10^{-1} \ \mathring{A}$,we multiply by $10/10$:$r = 17.32 \times 10^{-1} \ \mathring{A}$.Rounding to the nearest integer,we get $17 \times 10^{-1} \ \mathring{A}$.

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

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Sodium metal crystallizes in a body-centered cubic $(BCC)$ lattice with a unit cell edge length of $4 \ \mathring{A}$. The radius of the sodium atom is $..... \times 10^{-1} \ \mathring{A}$ (Nearest integer).

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A
$14$
B
$15$
C
$16$
D
$17$

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Mathematical analysis of cubic system and Bragg’s equation

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In a $CsCl$ crystal,the interionic distance between $Cs^+$ and $Cl^-$ ions is:

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Calculate the molar mass of a metal having a density of $22.24 \ g \ cm^{-3}$,which crystallizes to form a unit cell containing $4$ particles. Given $a^3 = 5.6 \times 10^{-23} \ cm^3$.

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Sodium metal crystallizes in a body-centered cubic $(BCC)$ lattice with a unit cell edge length of $4 \ \mathring{A}$. The radius of the sodium atom is $..... \times 10^{-1} \ \mathring{A}$ (Nearest integer).

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:

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?

Lithium has a $bcc$ structure. Its density is $530 \ kg \ m^{-3}$ and its atomic mass is $6.94 \ g \ mol^{-1}$. Calculate the edge length of a unit cell of lithium metal in $pm$ $(N_A = 6.02 \times 10^{23} \ mol^{-1})$.

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Calculate the molar mass of a metal having a density of $22.24 \ g \ cm^{-3}$,which crystallizes to form a unit cell containing $4$ particles. Given $a^3 = 5.6 \times 10^{-23} \ cm^3$.

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