Solid State - Result Question 39
####39. An element crystallises in fcc lattice having edge length $400 pm$. Calculate the maximum diameter of atom which can be placed in interstitial site without distorting the structure.
(2005, 2M)
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Answer:
Correct Answer: 39. $(117 pm)$
Solution:
- In a cubic crystal system, there are two types of voids known as octahedral and tetrahedral voids. If $r _1$ is the radius of void and $r _2$ is the radius of atom creating these voids then
$$ \left(\frac{r _1}{r _2}\right) _{\text {octa }}=0.414 \text { and }\left(\frac{r _1}{r _2}\right) _{\text {tetra }}=0.225 $$
The above radius ratio values indicate that octahedral void has larger radius, hence for maximum diameter of atom to be present in interstitial space :
$$ \begin{aligned} r _1 & =0.414 r _2 \\ \text { Also in fcc, } & 4 r _2=\sqrt{2} a \\ \Rightarrow \text { Diameter required }\left(2 r _1\right) & =\left(2 r _2\right) \times 0.414 \\ = & \frac{a}{\sqrt{2}} \times 0.414 \\ = & \frac{400 \times 0.414}{\sqrt{2}}=117 pm \end{aligned} $$
