Calculate the volume of a bcc unit cell if th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$7.951 	imes 10^{-23} \ cm^3$

Vedclass · Answer

(C) For a $bcc$ (body-centered cubic) unit cell,the relationship between the edge length $(a)$ and the atomic radius $(r)$ is given by: $4r = \sqrt{3}a$. Therefore,$a = \frac{4r}{\sqrt{3}}$. Given $r = 1.86 	imes 10^{-8} \ cm$. $a = \frac{4 	imes 1.86 	imes 10^{-8}}{\sqrt{3}} \approx \frac{7.44 	imes 10^{-8}}{1.732} \approx 4.2956 	imes 10^{-8} \ cm$. The volume of the unit cell $(V)$ is $a^3$. $V = (4.2956 	imes 10^{-8} \ cm)^3 \approx 79.28 	imes 10^{-24} \ cm^3 = 7.928 	imes 10^{-23} \ cm^3$. Rounding to the nearest option,the correct answer is $7.951 	imes 10^{-23} \ cm^3$.

Calculate the volume of a $bcc$ unit cell if the radius of an atom present in it is $1.86 \times 10^{-8} \ cm$.

Similar Questions

Calculate the molar mass of an element having a density of $8.6 \ g \ cm^{-3}$ if it forms a $bcc$ structure $[a^3 \times N_{A} = 22.0 \ cm^3 \ mol^{-1}]$.

What is the molar mass of a metal having a density of $8.57 \ g \ cm^{-3}$ and an edge length of $3.3 \ \mathring{A}$? (Packing efficiency $= 68 \%$)

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 ............. $\mathring{A}$.

$A$ metal crystallises in a body-centred cubic $(BCC)$ lattice with the metallic radius $\sqrt{3} \ \mathring{A}$. The volume of the unit cell in $m^3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module