Question

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure.The ratio of unit cells in 1000g at room temperature and above 900 degree celcius is (assuming molar mass & atomic radii of iron remains constant with temperature)

Answer 1

Answer:

Density of unit cell

d=Z×MNA×a3

where, Z=Number of atoms per unit cell

M=Molar mass

a3=Volume of unit cell [a=edge length]

NA=Avogadro's number =6.022×1023

For bcc, Z=2, radius (r )=3–√a4

a=4r3–√

For fcc, Z=4,r=a22–√

⇒a=22–√r

According to question

droom temp.D900∘C=(ZMNAa3)bcc(ZMNAa3)fcc

On substituting the given values , we get

droom temp.d900∘C=2×MNA×(4r3√)3/4×MNA×(22–√r)3

[∵ Given, M and r of iron remains constant with temperature]

=2×33–√64r3×162–√r34

dbccdfcc=3432−−√

Explanation: