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Copper has face centered cubic (fcc) lattice with interatomic spacing equal to 2.54 Å. The value of lattice constant for this lattice is -
A
3.59 Å
B
2.54 Å
C
1.27 Å
D
5.08 Å
Detailed Solution
In an fcc lattice the atoms touch each other along the face diagonal: $4r = \sqrt{2}\,a$
The interatomic spacing (nearest neighbour distance) is $2r$.
So $2r = \dfrac{\sqrt{2}\,a}{2} = \dfrac{a}{\sqrt{2}}$
Given $2r = 2.54$ Å, so $a = \sqrt{2} \times 2.54$
$a = 1.414 \times 2.54$
$a = 3.59$ Å
The interatomic spacing (nearest neighbour distance) is $2r$.
So $2r = \dfrac{\sqrt{2}\,a}{2} = \dfrac{a}{\sqrt{2}}$
Given $2r = 2.54$ Å, so $a = \sqrt{2} \times 2.54$
$a = 1.414 \times 2.54$
$a = 3.59$ Å