Each edge of a cubic unit cell is 400 pm long. If atomic mass of the element is 120 and its density is 6 ⋅ 25 g / cm 3, the crystal lattice is: (use NA =6 × 1023 )

Question

Each edge of a cubic unit cell is 400 pm long. If atomic mass of the element is 120 and its density is 6 ⋅ 25 g / cm 3, the crystal lattice is: (use NA =6 × 1023 ) (A) primitive (B) body centered (C) face centered (D) end centered

Tardigrade Admin · Accepted Answer

d =6.25 g / cm 3 M =120 d a =400 pm =4 × 10-8 cm We know, d=(2 × M/a3 × NA) ∴ 6.25 =z × 120 =(2(4 × 10-8)3 × 6.022 × 1023/2)=2 ∴ It is body centered

Q. Each edge of a cubic unit cell is $400 p m$ long. If atomic mass of the element is $120$ and its density is $6 \cdot 25 g / c m^{3}$ , the crystal lattice is : (use $N_{A} = 6 \times 1 0^{23}$ )

primitive

body centered

face centered

end centered

$d = 6.25 g / c m^{3}$
$M = 120 d$
$a = 400 p m$
$= 4 \times 1 0^{- 8} c m$
We know, $d = \frac{2 \times M}{a ^{3} \times N _{A}}$
$∴ 6.25 = z \times 120$
$= \frac{2 ( 4 \times 1 0 ^{- 8} ) ^{3} \times 6.022 \times 1 0 ^{23}}{2} = 2$
$∴$ It is body centered

Q. Each edge of a cubic unit cell is 400pm long. If atomic mass of the element is 120 and its density is 6⋅25g/cm3, the crystal lattice is : (use NA​=6×1023 )