Question

Calculate the atomic radius of elementary silver which crystallizes in fcc lattice with unit cell edge length 4.086X10^-10

Answer 1

Answer:

1.444X10^-10

Explanation:

in fcc we know that 4r=2^1/2(a)

therefore 4r=2^1/2(4.086(10^-10))

r=1.444X10^-10

Answer 2

Answer:

1.44 × 10^-10 m

Explanation:

a= 4.086×10^-10 m

We know, In fcc if 'r' is the radius and 'a' is the edge length, the the relationship between r and a is

a= 2√2 r

r= a/2√2

=4.086×10^-10/ 2× 1.414 [since √2= 1.414]

=4.086/2.828 ×10^-10

r = 1.44 × 10^-10 Ans