UPSEE Solved Model Paper 7

According to the question

Fe

(bcc)

>916°C

→

Fe

(ccp)

where, bcc = Body centred cube
ccp = Cubic dose packed structure
The density of unit cell is given as
ρ =

Z×M

a3×N0

... (i)
where, 2 = Number of atoms per unit cell
M = Atomic mass
a = Edge length
No = Avogadro's constant = 6 023 × 1023mol−1
The density of iron (bcc structure) before heating
For a bcc structure, Z = 2
Also,
a =

4r

√3

where, r = Radius of the sphere
Substituting the values of Z and a in equation (i). we get
ρ =

2M

( 4r √3 )3×N0

... (ii)
The density of iron (ccp structure) after heating is For a ccp structure. Z = 4
Also,
a = 2√2 r (Since radius of atom remains the same)
Substituting the values of T and a' in equation (i), we get
ρ' =

4M

(2√2r)3×N0

... (iii)
Dividing equation (I) by equation (ii), we get
The ratio of density

ρ

ρ′

= 0.918
Therefore, the ratio of density of the crystal before heating and after heating is 0.918
Hence, option 'B' is correct