Question

Calculate the number of atoms in a critically sized nucleus for the homogeneous nucleation of pure lead. Assume 20%Tm under-cooling.

Answer 1

we know that Pb has

Hf= heat of fusion= 280 J/cm^3

y= energy of surface= 33.3 x10^-7 J/cm^2

so ,Tm= 80°C = 353.15 K

the critically radius is:

r= (2x(33.3 x10^-7)x353.15)/(280x 0.2 x 353.15)

= 1.18929x10^-7 cm

Volume critically = (4/3)(pi)r^3

= 7.04607x10^-21 cm^3

the radius of Pb is a=495x10^-10 cm

so a^3 of a unit atom is a^3= 1.22x10^-22 cm^3

in a BCC are 2 atoms so

Volume/ atom= 1.22x10^-22 / 2

= 6.06x10^-23 cm^3

so, atoms= vol nucleous/ vol / atom

= 7.04607x10^-21 / 6.06x10^-23

= 116.187 atoms