Consider homogeneous nucleation of a spherical solid in liquid. For a given undercooling, if surface energy of a nucleus increases by $20\%$, the corresponding increase (in percent) in the critical radius of the nucleus is: ___________(round off to nearest integer).
Understanding Critical Radius in Nucleation
The critical radius ($r_c$) is the minimum radius a nucleus must achieve to grow stably. For homogeneous nucleation in a spherical solid forming from a liquid, the critical radius depends on the interfacial energy ($\gamma$) and the undercooling ($\Delta T$).
Nucleation Formula and Variables
The formula for the critical radius ($r_c$) is:
$r_c = \frac{2 \gamma T_m}{\Delta H_f \Delta T}$
- $\gamma$: Interfacial energy between solid and liquid.
- $T_m$: Melting temperature.
- $\Delta H_f$: Latent heat of fusion.
- $\Delta T$: Undercooling (temperature difference from melting point).
Calculating the Effect of Increased Surface Energy
The question states that the undercooling ($\Delta T$) is constant. The surface energy ($\gamma$) increases by $20\%$. Let the initial surface energy be $\gamma_1$ and the final surface energy be $\gamma_2$.
Then, $\gamma_2 = \gamma_1 + 0.20 \gamma_1 = 1.20 \gamma_1$.
Let the initial critical radius be $r_{c1}$ and the final critical radius be $r_{c2}$.
$r_{c1} = \frac{2 \gamma_1 T_m}{\Delta H_f \Delta T}$
$r_{c2} = \frac{2 \gamma_2 T_m}{\Delta H_f \Delta T} = \frac{2 (1.20 \gamma_1) T_m}{\Delta H_f \Delta T}$
Substituting the expression for $r_{c1}$ into the equation for $r_{c2}$:
$r_{c2} = 1.20 \times \left( \frac{2 \gamma_1 T_m}{\Delta H_f \Delta T} \right) = 1.20 r_{c1}$
Determining the Percentage Increase
The increase in the critical radius is $\Delta r_c = r_{c2} - r_{c1} = 1.20 r_{c1} - r_{c1} = 0.20 r_{c1}$.
The percentage increase is calculated as:
$\text{Percentage Increase} = \frac{\Delta r_c}{r_{c1}} \times 100\% = \frac{0.20 r_{c1}}{r_{c1}} \times 100\%$
$\text{Percentage Increase} = 0.20 \times 100\% = 20\%$
Rounding to the nearest integer, the increase is $20\%$. Since the critical radius ($r_c$) is directly proportional to the surface energy ($\gamma$) when undercooling ($\Delta T$) is constant, a $20\%$ increase in $\gamma$ results in a $20\%$ increase in $r_c$.
Important Questions from Solidification Cooling Curve Analysis
- Critical value of the Gibbs energy of nucleation at equilibrium temperature is
- During the solidification of a pure metal, it was found that dendrites are formed. Assuming that the liquid-solid interface is at the melting temperature, the temperature from the interface into the liquid
- Which one of the following schematics represents the variation of the rate of nucleation of solid from a pure liquid metal as a function of undercooling ($\Delta T = T_m - T$, where $T_m$ and $T$ are the freezing temperature and the liquid temperature, respectively)?
A given volume of liquid is undercooled just below the melting temperature to form a spherical solid nucleus (consider homogeneous nucleation). The Gibbs free energy of solidification ($\Delta G_v$) is ($- 0.5 \times 10^8$) J/m$^3$. The solid-liquid interfacial energy ($\gamma$) is isotropic and its value is 0.1 J/m$^2$.
The critical nucleus size for a stable nucleus is __________ nm (answer in integer).
During solidification of a pure metal, the radius of critical nucleus at an undercooling of 10 K is ________ $\times 10^{-9} \text{ m}$ (answer rounded off to 1 decimal place).
Given: solid/liquid interface energy = $0.177 \text{ J} \cdot \text{m}^{-2}$,
melting point of the metal = 1356 K and
latent heat of fusion = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$
