Question

Assume the radius of an atom, which can be represented as a hard sphere, is r  1.95 Å.The atom is placed in a ( a ) simple cubic, ( b ) fcc, ( c ) bcc, and ( d ) diamond lattice. As-suming that nearest atoms are touching each other, what is the lattice constant of eachlattice

Answer 1

Answer: Simple cubic=0.39nm

Face centred cubic

=0.55nm

Body centered cubic

=0.45nm

Diamond lattice= 0.9nm

Step-by-step explanation: The lattice constant (a)

for SC=2*r

Fcc=4*r/√2

Bcc= 4*r/√3

Diamond lattice=8*r/√3

Here,

r is the atomic radius measured in nm

r = 1.95Å * 1nm/10Å

=0.195nm

Now let's calculate (a)

SC = 2*r = 2*0.195 nm=0.39nm

Fcc = 4*r/√2 =4*0.195nm/√2

= 0.55nm

Bcc = 4*r/√3 =4*0.195nm/√3

= 0.45nm

Diamond lattice = 8*r/√3

=8*0.195nm/√3

= 0.9nm