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(\dfrac{r _1}{r _2}\right) _{\text {octa }}=0.414 \text { and }\left(\dfrac{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 :
$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 $
$= \dfrac{a}{\sqrt{2}} \times 0.414 $
$= \dfrac{400 \times 0.414}{\sqrt{2}}=117$ $ pm$
BETA
