Iron exhibits bcc structure at room temperature. Above 500^ C , it transforms to fcc structure. Find the ratio of the density of iron at room temperature to that at 500^ C . (Assume the atomic radii and the molar mass of iron remain constant even with variation in temperature)

Step-by-step solution

We know, Density, d=Z M / a^3 N₀ For bcc unit cell, Z=2, a= 4 3 r d_ bcc = 2 M ( 4 3 r )^3 N₀ For fcc unit cell, Z=4, a= 4 2 r d_ fcc = 4 M ( 4 2 r )^3 N₀ Ratio of density of iron bcc : fcc can be calculated by dividing Eq. (i) by Eq. (ii) d_ bcc d_ fcc = 2 M ( 4 3 r )^3 N₀ 4 M ( 4 2 r )^3 N₀ = 1 2 ( 3 2 )^3 = 1 2 3 2 3 2 = 3 3 4 2 The ratio is 3 3 : 4 2 .