Solid State - Result Question 38
####38. The edge length of unit cell of a metal having molecular weight $75 g / mol$ is $5 \AA$ which crystallises in cubic lattice. If the density is $2 g / cc$ then find the radius of metal atom. $\left(N _A=6 \times 10^{23}\right)$. Give the answer in pm.
(2006, 3M)
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Answer:
Correct Answer: 38. $(217 pm)$
Solution:
- From the given information, the number of atoms per unit cell and therefore, type of unit cell can be known as
$$ \begin{aligned} \rho & =\frac{N M}{N _A a^{3}} \\ \Rightarrow \quad N & =\frac{\rho N _A a^{3}}{M}=\frac{2 \times 6 \times 10^{23} \times\left(5 \times 10^{-8} cm\right)^{3}}{75}=2(bcc) \\ \Rightarrow \quad \text { In bcc, } 4 r & =\sqrt{3} a \\ \Rightarrow \quad r & =\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 5 \times 10^{-10} m \\ & =2.17 \times 10^{-10} m=217 pm \end{aligned} $$
