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  • An element has a body-centred cubic (bcc) structure with a cell edge of 288pm. The atomic radius is:Option: 1 <img alt="\frac{4}{\sqrt{2}}\times

An element has a body-centred cubic (bcc) structure with a cell edge of 288pm. The atomic radius is:

Option: 1

\frac{4}{\sqrt{2}}\times 288\; \text {pm}


Option: 2

\frac{\sqrt{3}}{4}\times 288\; \text {pm}


Option: 3

\frac{\sqrt{2}}{4}\times 288\; \text {pm}


Option: 4

\frac{4}{\sqrt{3}}\times 288\; \text {pm}


Answers (1)

best_answer

We know the relation between r and a for the BCC unit cell

\mathrm{4r = \sqrt{3}a}

So,

\mathrm{r = \frac{\sqrt{3}}{4}a}

\mathrm{r = \frac{\sqrt{3}}{4}\times 288 \ pm}

Therefore, the correct option is (2).

Posted by

manish

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