2020 The Solid StateCalculations Involving Unit Cell Dimensions Easy Question 11 of 45 in The Solid State

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

A $\frac{\sqrt{2}}{4}\times288\ pm$
B $\frac{4}{\sqrt{3}}\times288\ pm$
C $\frac{4}{\sqrt{2}}\times288\ pm$
D $\frac{\sqrt{3}}{4}\times288\ pm$

The answer is D.

Solution

For bcc: $\sqrt{3}a = 4r \Rightarrow r = \frac{\sqrt{3}}{4}a$ With a = 288 pm: $r = \frac{\sqrt{3}}{4} \times 288$ pm

Concept atomic radius in a bcc lattice

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45 questions from this chapter have appeared across 23 exam years.