Given :
Edge length of a unit cell = 4.53
= 4.53 ×
10−8 cm
Molecular weight = 24
Density, d = 1.74
g∕cm3 Avogadro's number = 6 ×
1023 We have to find the radius of the metal.
Now,
Density is given as:
ρ =
| z×Molecularweight |
| a3×Avogadro′snumber |
... (i)
where, a = Edge length
z = No. of atoms in a unit cell
From equation (i):
z =
| ρ×a3×Avogadro′snumber |
| Molecularweight |
... (ii)
On substituting all the values in equation (ii), we get
z =
z =
z =
z = 4.043
z ~ 4
As, the value of 'z' is 4. thus, metal crystallises in face centred cubic (fcc) lattice
Now.
For fcc structure, radius is related to edge length as
r =
... (iii)
Substituting all the values in equation (iii), we get
r =
cm
r =
cm
r = 1.60 ×
10−8 cm
or
r = 160 ×
10−10 cm
or
r = 160 ×
10−12 m
r = 160 pm
Hence, option 'B' is correct