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Chemistry Question on Unit Cells

If a metal crystallises in bcc structure with edge length of unit cell 4.29×10−8 cm4.29 \times 10^{-8}\, cm the radius of metal atom is

A

3.2×10−7 cm3.2 \times 10^{-7}\,cm

B

1.86×10−8 cm1.86 \times 10^{-8}\,cm

C

1.07×10−7 cm1.07 \times 10^{-7}\,cm

D

1.07×10−8 cm1.07 \times 10^{-8}\,cm

Answer

1.86×10−8 cm1.86 \times 10^{-8}\,cm

Explanation

Solution

Key Idea The relation between edge length and radius of an atom in a bccbcc structure is related as:

r=3/4ar=\sqrt{3} / 4 a

Given, a=4.29×10−8cma=4.29 \times 10^{-8} cm

For bcc structure,

r=34a=34×4.29×10−8cmr =\frac{\sqrt{3}}{4} a=\frac{\sqrt{3}}{4} \times 4.29 \times 10^{-8} cm
=1.857×10−8cm≃186×10−8cm=1.857 \times 10^{-8} cm \simeq 186 \times 10^{-8} cm