Question

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) 2.54 Å (b) 3.59 Å
(c) 1.27 Å (d) 5.08 Å

Answer 1

(c) 1.27 Å

I hope it helps you

Answer 2

Answer:

Copper has a face-centered cubic (fcc) lattice with interatomic  spacing equal to 2.54 Å. The value of the lattice constant for  this lattice is

Explanation:

The d- spacing or interatomic spacing of a cubic lattice is the distance between the consecutive lattice planes of a lattice. Which  is given by,

$d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$  ..(1)

Where,

h,k,l    -  Miller indices

a         - lattice constant

We have the expression for atomic spacing as,

$d = 2 r=2.54 \ A^{\circ}$  ..(2)

For face-centered cubic lattices (FCC lattice),

$4r = \sqrt{2} a$

Dividing throughout the equation by $2$

$\frac{4r}{2} =\frac{ \sqrt{2} a}{2}$

or,

$2r = \frac{a}{\sqrt{2} }$      ...(3)

Using equations(2) and (3),

$\frac{a}{\sqrt{2} } = 2.54\ A^{\circ}$

Then,

a  $= 2.54\times {\sqrt{2}$

   = $\left(2.54 \mathrm{~A}^{\circ}\right)(1.414)=3.59 \mathrm{~A}^{\circ}$

Thus option b is the correct answer.