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Activation free energy for nucleation

  1. Jan 1, 2009 #1
    Hi everyone,


    In metals, for homogeneous nucleation, activation free energy required for the formation of a stable nucleus are different when the nucleus are considered as a cube and considered as a sphere and the relation between them is energy for cube is almost double of the energy for the sphere. Why?

    The answer I gave for this question is;

    During the transformation, the total energy is used to create the surface, since the atoms on the surface aren't in equilibrium. For sphere and cube which have the same volume, the surface of the cube is greater than the other. Thus, more energy is needed to create a stable nucleus when its shape is cube.

    This was our quiz question and I got zerofrom that! However, I still think that this answer is correct. Could you please tell me the faults and missimg parts that you think could be the answer?

    Thank you.
  2. jcsd
  3. Jan 1, 2009 #2


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    The question isn't just asking why the activation free energy is larger, it's asking why it's almost twice as large (actually, 1.9 times). Your answer attributes this increase to surface area, but the surface area of a cube is only 24% larger than that of a sphere of identical volume. I trust that's why your answer was marked wrong.

    A better answer would have been to show how the activation free energy is calculated (by differentiating the total volumetric and surface area energy with respect to cluster size). This approach yields the 1.9x figure.
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