Question

calculate the fermi energy of face centered cubic crystal of silver of atomic radius 1.44A​

Answer 1

Answer:

Given,

Atomic mass of silver =108u

Atomic cell length, a=4.077×10

−8

cm

Atomic radius in FCC, r=

4

a

2

=

4

4.077×10

−8

2

=1.44×10

−8

cm\

Each silver unit cell contains 4 atom.

Density =

N

A

×a

3

4×atomicmassofsilver

=

6.02×10

23

×(4.077×10

−10

)

3

4×108

=10589.25kgm

−3

Hence, atomic radius 1.44×10

−8

cm and density is 10589.25kgm

−3