A spherical $\beta$ particle nucleates from the $\alpha$ matrix on a non-deformable substrate $\delta$, forming a contact angle $\theta$ as shown. The value of $\dfrac{\Delta G^{}_{het}}{\Delta G^{}_{hom}}$ is _________.
(Round off to three decimal places)
Given:
$\alpha$–$\beta$ interfacial energy $= 0.4$ J/m$^{2}$
$\alpha$–$\delta$ interfacial energy $= 0.3$ J/m$^{2}$
$\beta$–$\delta$ interfacial energy $= 0.02$ J/m$^{2}$
Show Hint
Use Young’s equation to get the contact angle, then apply the spherical-cap reduction factor $f(\theta)$ for heterogeneous nucleation.
For heterogeneous nucleation of a spherical cap on a flat substrate, the ratio of heterogeneous to homogeneous nucleation energy is:
\[
\frac{\Delta G^{}_{het}}{\Delta G^{}_{hom}}
= f(\theta)
\]
where
\[
f(\theta)= \frac{1}{4}(2+\cos\theta)(1-\cos\theta)^2
\]
The contact angle is obtained from Young’s equation:
\[
\gamma_{\alpha\delta} = \gamma_{\alpha\beta} \cos\theta + \gamma_{\beta\delta}
\]
Substitute values:
\[
0.3 = 0.4\cos\theta + 0.02
\]
\[
0.28 = 0.4 \cos\theta
\]
\[
\cos\theta = 0.70
\]
Now compute the shape factor:
\[
f(\theta)= \frac{1}{4}(2+\cos\theta)(1-\cos\theta)^2
\]
\[
f(\theta)= \frac{1}{4}(2+0.7)(1-0.7)^2
\]
\[
f(\theta)= \frac{1}{4}(2.7)(0.3^2)
\]
\[
f(\theta)= \frac{1}{4}(2.7)(0.09)
\]
\[
f(\theta)= 0.06075
\]
Rounded to three decimals:
\[
\boxed{0.061}
\]
