= 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 =
1.74×(4.53)3×(10−8)3×6×1023
24
z =
1.74×92.96×6×10−24×1023
24
z =
97.05
24
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 =
a
2√2
... (iii) Substituting all the values in equation (iii), we get r =
4.53×10−8
2×1.414
cm r =
4.53×10−8
2.828
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