Question

Calculate the critical radius (in centimeter) of a homogeneous nucleus that forms when pure liquid copper solidifies. Assume Delta T (under cooling) =0.2T_m. Where T_m=1083 degree C. Values for the latent heat of fusion and surface free energy are 1826 J/cm^3 and 177 times 10^-7 J/cm^2, respectively. Calculate the number of atoms in the critical sized nucleus at this under cooling. Copper has FCC crystal structure. Assume a lattice parameter of 0.361 nm for solid copper at its melting temperature

Answer 1

a. $\small \Delta T = 0.2T_{m}$

crtical radius,

$\small r^{*} = \frac{2\sigma _{sl}T_{m}}{\Delta H_{f}\Delta T}$

substituting all the given values,

$\small r^{*} = \frac{2\times 177\times 10^{-7}\times1083 }{1826\times 216.6}$

$\small r^{*} = 9.693 \times 10^{-8} cm$

b. lattice parameter, a = 0.361 nm = 3.61 x 10^-8 cm

for FCC, $\small r = \frac{a\sqrt{2}}{4}$

$\small r = \frac{3.61\times 10^{-8}\times \sqrt{2}}{4}$

$\small r = 1.276 \times 10^{-8} cm$

number of unit cell = r*/r = 7.596

number of atoms per unit cell of FCC = 4

number of atoms = 30.386