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 = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$
Critical Nucleus Radius Calculation
The radius of the critical nucleus ($r^*$) during solidification is determined by the interface energy ($\gamma$), melting point ($T_m$), latent heat of fusion ($L_f$), and undercooling ($\Delta T$).
Formula for Critical Nucleus Radius
The formula used is:
$r^* = \frac{2 \gamma T_m}{L_f \Delta T}$
Given Parameters
- Solid/liquid interface energy ($\gamma$): $0.177 \text{ J} \cdot \text{m}^{-2}$
- Melting point ($T_m$): $1356 \text{ K}$
- Latent heat of fusion ($L_f$): $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$
- Undercooling ($\Delta T$): $10 \text{ K}$
Step-by-Step Calculation
- Substitute the given values into the formula: $r^* = \frac{2 \times 0.177 \text{ J/m}^2 \times 1356 \text{ K}}{(1.88 \times 10^9 \text{ J/m}^3) \times 10 \text{ K}}$
- Calculate the numerator: $2 \times 0.177 \times 1356 = 480.264 \text{ J/m}^2 \cdot \text{K}$
- Calculate the denominator: $(1.88 \times 10^9) \times 10 = 1.88 \times 10^{10} \text{ J/m}^3 \cdot \text{K}$
- Compute the radius: $r^* = \frac{480.264 \text{ J/m}^2 \cdot \text{K}}{1.88 \times 10^{10} \text{ J/m}^3 \cdot \text{K}} = 2.554595 \times 10^{-8} \text{ m}$
- Convert the result to the required units ($\times 10^{-9} \text{ m}$): $r^* = 2.554595 \times 10^{-8} \text{ m} = 25.54595 \times 10^{-9} \text{ m}$
- Round off to one decimal place as requested: $r^* \approx 25.5 \times 10^{-9} \text{ m}$
Result Verification
The calculated value of $25.5 \times 10^{-9} \text{ m}$ falls within the given correct answer range of 25.1 to 25.9 ($\times 10^{-9} \text{ m}$).
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).
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).
