Question

Unit cell of iron crystal has edge length of 288 pm and density of 7.86 g cm⁻³. Determine the type of crystal lattice (Fe=56). (Hint : Determine number of atoms in unit cell, it comes to 2, hence bcc type)

Answer 1

Hey Dear,

● Answer -

BCC crystal lattice

● Explanation -

# Given -

a = 288 pm = 288×10^-10 cm

d = 7.86 g/cm³

M = 56 g/mol

N = 6.022×10^23 particles/mol

# Solution -

Density of the unit cell is calculated by -

d = Z.M / N.a³

Z = d × N × a³ / M

Z = 7.86 × 6.022×10^23 × (288×10^-10)^3 / 56

Z = 2

Here, presence of 2 atoms per unit cell indicates BCC lattice.

Hope this helps you.

Answer 2

Answer:

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