Question

determine the lattice constant for FCC lead crystal of radius
1.746amstrong​

Answer 1

Answer:

hope it helps .

Explanation:

Given:

It is an FCC structure

r = 1,746 A

Plane (2 2 0)

To find: a = ? and d = ?

Solution:

In FCC structure,

a=4r2√=4×1746×10−102√

a=4938.43×10−10m

d=a22+22+02√

d=4938.43×10−108√

d=1745.99×10−10m