MHT CET · Chemistry · Solid State
An element has BCC structure with edge length of unit cell 600 \(\mathrm{pm}\). What is the atomic radius of element?
- A \(\sqrt{3} \times 150 \mathrm{pm}\)
- B \(150 \mathrm{pm}\)
- C \(300 \mathrm{pm}\)
- D \(\frac{300}{\sqrt{2}} \mathrm{pm}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{3} \times 150 \mathrm{pm}\)
Step-by-step Solution
Detailed explanation
For BCC structure.
\(\begin{aligned}
& \sqrt{3} \mathrm{a}=4 \mathrm{r} \\
& \mathrm{r}=\frac{\sqrt{3} \mathrm{a}}{4}=\frac{\sqrt{3} \times 600}{4} \\
& \mathrm{r}=\sqrt{3} \times 150 \mathrm{pm}
\end{aligned}\)
\(\begin{aligned}
& \sqrt{3} \mathrm{a}=4 \mathrm{r} \\
& \mathrm{r}=\frac{\sqrt{3} \mathrm{a}}{4}=\frac{\sqrt{3} \times 600}{4} \\
& \mathrm{r}=\sqrt{3} \times 150 \mathrm{pm}
\end{aligned}\)
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