# What exactly is nucleation

## Heterogeneous nucleation

In heterogeneous nucleation, nuclei that do not consist of the same substance as the melt trigger solidification.

### introduction

The nucleation considered in the section homogeneous nucleation referred to the own particles in the melt (own nuclei). The greatest inhibition threshold for this homogeneous nucleation is the application of surface energy. This means that small germs are unable to grow and accordingly dissolve.

A reduction in surface energy therefore means that a lower activation energy is required for nucleation. This means that even smaller germs are stable and do not dissolve again immediately. The thermal fluctuations would then be increasingly able to form germs capable of growth.

At this point, contamination, foreign particles or the vessel walls can be very helpful (generally also as *Substrate* because it is precisely these that provide part of the surface energy required for nucleation. One then speaks of *heterogeneous nucleation*which requires less activation work (nucleation work) than homogeneous nucleation.

Foreign particles can do part of the nucleation work, so that even smaller nuclei are able to grow!

Since a melt usually always contains such foreign bodies, the heterogeneous nucleation is therefore much more likely than the homogeneous nucleation.

### Wetting

The derivation of the laws for the heterogeneous nucleation takes place in a similar way as for the homogeneous nucleation. As an example, a smooth vessel wall (mold) is intended to serve towards the melt in the following, on which a nucleus is formed. Due to the surface tension that has been applied, the solidified nucleus has the shape of a cut-off sphere.

The angle at the point of contact between the surface of the germ and the wall of the vessel is also called *Wetting angle* \ (\ Theta \) or *Contact angle* designated. Depending on the interaction between the nucleus, melt and wall, the wall is very heavily wetted by the nucleus (low wetting angle) or hardly wetted (large wetting angle). This accordingly has an impact on the shape of the spherical segment-shaped nucleus.

The decisive factors for the contact angle \ (\ Theta \) are *specific surface energies* or. *Surface tension* \ (\ gamma \) between the said interfaces germ-melt (\ (\ gamma_ {KS} \)), germ-wall (\ (\ gamma_ {KW} \)) and melt-wall (\ (\ gamma_ {SW } \)). An equilibrium consideration results in the following relationship (*Young equation*):

\ begin {align}

\ label {wetting angle}

\ boxed {\ cos \ Theta = \ frac {\ gamma_ {SW} - \ gamma_ {KW}} {\ gamma_ {KS}}} ~~~ \ text {Young equation} \ [5px]

\ end {align}

On the basis of the wetting angle \ (\ theta \), the volume \ (V \) of the spherical segment for a given nucleus radius \ (r \) can be determined using the equation given below:

\ begin {align}

& V = \ tfrac {1} {3} h ^ 2 \ pi \ left (3r - h \ right) ~~~ \ text {with} ~~~ h = r \ left (1- \ cos \ Theta \ right) ~~~ \ text {follows:} \ [5px]

\ label {coupling volume}

& \ boxed {V = \ tfrac {1} {3} \ pi \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) r ^ 3} \ [5px]

\ end {align}

The surface of the spherical cap \ (O_S \), which forms the interface between the nucleus and the melt, can also be determined via the wetting angle \ (\ Theta \).

\ begin {align}

& O_S = 2 \ pi r h ~~~ \ text {with} ~~~ h = r \ left (1- \ cos \ Theta \ right) ~~~ \ text {follows:} \ [5px]

\ label {jacket area}

& \ boxed {O_S = 2 \ pi \ left (1- \ cos \ Theta \ right) r ^ 2} \ [5px]

\ end {align}

The base area \ (O_W \) of the spherical segment that forms towards the wall is also dependent on the wetting angle \ (\ Theta \):

\ begin {align}

& O_W = \ pi a ^ 2 ~~~ \ text {with} ~~~ a = r ~ \ sin \ Theta ~~~ \ text {follows:} \ [5px]

\ label {base area}

& \ boxed {O_W = \ pi \ sin ^ 2 \ Theta \ cdot r ^ 2} \ [5px]

\ end {align}

### Gibbs energy of a germ

In the case of heterogeneous nucleation, the change in Gibbs energy \ (\ Delta G_ {nucleus} \) of a nucleus that forms is made up of three parts:

\ begin {align}

\ label {0}

\ boxed {\ Delta G_ {Keim} = \ Delta G_ {V} + \ Delta G_ {O, KS} + \ Delta G_ {O, KW}}

\ end {align}

- Reduction of the volume energy \ (\ Delta G_V \) due to the phase change
- Surface energy to be applied \ (\ Delta G_ {O, KS} \) at the interface between nucleus and melt
- Change of the surface energy \ (\ Delta G_ {O, KW} \) during the phase change between the nucleus and the wall

These terms are discussed in more detail below.

### Reduction in volume energy due to the phase change

The change in the "volume energy" \ (\ Delta G_V \) is determined by the volume \ (V \) of the nucleus in combination with the density \ (\ rho \) and the specific Gibbs energy change \ (\ Delta g_ {V} \ ) given (note that this part leads to a lowering of the energy and therefore has a negative sign in contrast to the surface energy to be applied):

\ begin {align}

& \ Delta G_ {V} = - m \ cdot \ Delta g_V ~~~~~ \ text {with} ~~~~ m = V \ cdot \ rho ~~~~ \ text {follows:} \ [5px ]

\ label {321}

& \ underline {\ Delta G_ {V} = - V \ cdot \ rho \ cdot \ Delta g_V} \ [5px]

\ end {align}

The specific change in Gibbs energy \ (\ Delta g_ {V} \) can be expressed by the relation to the specific heat of fusion \ (q_E \), solidification temperature \ (T_E \) and to supercooling \ (\ Delta T \) (derivation look here):

\ begin {align}

\ label {melt}

& \ underline {\ Delta g_V = q_E \ cdot \ frac {\ Delta T} {T_E}} \ [5px]

\ end {align}

If the equations (\ ref {dome volume}) and (\ ref {schmelz}) are inserted into equation (\ ref {321}), the volume energy \ (\ Delta G_V \) to be expended results from the phase change of the nucleus as follows:

\ begin {align}

\ label {2}

& \ boxed {\ Delta G_ {V} = - \ frac {\ pi ~ \ rho ~ q_E} {3 ~ T_E} \ Delta T \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) \ cdot r ^ 3} \ [5px]

\ end {align}

### Surface energy to be applied at the interface between nucleus and melt

The surface energy \ (\ Delta G_ {O, KS} \) to be used to generate the nucleus surface towards the melt results from the spherical cap \ (O_S \) and the corresponding specific surface energy \ (\ gamma_ {KS} \):

\ begin {align}

& \ underline {\ Delta G_ {O, KS} = \ gamma_ {KS} \ cdot O_S} \ [5px]

\ end {align}

With the help of equation (\ ref {mantelflaeche}) the surface energy \ (\ Delta G_ {O, KS} \) to be used to generate the nucleus surface towards the melt is given as follows:

\ begin {align}

\ label {3}

& \ boxed {\ Delta G_ {O, KS} = 2 \ pi ~ \ gamma_ {KS} ~ \ left (1- \ cos \ Theta \ right) r ^ 2} \ [5px]

\ end {align}

### Change in surface energy at the contact surface with the wall

In the molten state, the surface energy \ (E_ {SW} \) at the contact area between the not yet formed nucleus (= melt) and the wall results as follows:

\ begin {align}

E_ {SW} = \ gamma_ {SW} \ cdot O_W \ [5px]

\ end {align}

Due to the formation of the nucleus, the surface energy now changes between the former melt / wall to the nucleus / wall with the corresponding surface energy \ (E_ {KW} \):

\ begin {align}

E_ {KW} = \ gamma_ {KW} \ cdot O_W \ [5px]

\ end {align}

Consequently, due to the solidified germ, the surface energy at the contact point with the wall - and with it the Gibbs energy of the germ - changed as follows:

\ begin {align}

\ label {d}

& \ underline {\ Delta G_ {O, KW} = (\ gamma_ {KW} - \ gamma_ {SW}) \ cdot O_W} \ [5px]

\ end {align}

The energetic advantage of heterogeneous nucleation is already evident at this point. The surface energy of the germ towards the wall does not have to be applied from zero but only changed (\ (\ gamma_ {SW}> 0 \))!

While the contact surface \ (O_W \) is determined by the equation (\ ref {base surface}), the difference in surface tensions can be replaced by the equation (\ ref {wetting angle}):

\ begin {align}

\ label {young}

& \ underline {\ gamma_ {KW} - \ gamma_ {SW} = - \ gamma_ {KS} \ cdot \ cos (\ Theta)} \ [5px]

\ end {align}

If the equations (\ ref {base area}) and (\ ref {young}) are inserted into the equation (\ ref {d}), then the change in surface energy at the contact point to the wall can be \ (\ Delta G_ {O, KW } \) can be determined as follows:

\ begin {align}

\ underline {\ Delta G_ {O, KW} = - \ gamma_ {SW} \ cdot \ cos \ Theta \ cdot \ pi \ sin ^ 2 \ Theta \ cdot r ^ 2} \ [5px]

\ end {align}

Furthermore, it can be used that the geometric expression \ (\ cos (\ Theta) \ cdot \ sin ^ 2 (\ Theta) \) is replaced by the term \ (\ cos (\ Theta) - \ cos ^ 3 (\ Theta) \ ) can be replaced. Finally, the following applies:

\ begin {align}

\ label {1}

& \ boxed {\ Delta G_ {O, KW} = - \ pi ~ \ gamma_ {SW} ~ \ left (\ cos \ Theta - \ cos ^ 3 \ Theta \ right) \ cdot r ^ 2} \ [5px ]

\ end {align}

### Total change in the Gibbs energy of the nucleus

The change in the Gibbs energy \ (\ Delta G_ {Keim} \) of the solidifying nucleus can be calculated by inserting the equations (\ ref {2}), (\ ref {3}) and (\ ref {1}) into equation ( \ ref {0}) and then summarizing can be expressed as follows:

\ begin {align}

\ Delta G_ {Keim} = & - \ frac {\ pi ~ \ rho ~ q_E} {3 ~ T_E} ~ \ Delta T \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) \ cdot r ^ 3 \ [5px]

& + 2 \ pi ~ \ gamma_ {KS} ~ \ left (1- \ cos \ Theta \ right) r ^ 2 \ [5px]

& - \ pi ~ \ gamma_ {KS} ~ \ left (\ cos \ Theta - \ cos ^ 3 \ Theta \ right) \ cdot r ^ 2 \ [5px]

\ Delta G_ {Keim} = & - \ frac {\ pi ~ \ rho ~ q_E} {3 ~ T_E} ~ \ Delta T \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) \ cdot r ^ 3 \ [5px]

& + \ pi ~ \ gamma_ {KS} ~ \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) ~ r ^ 2 \ [5px]

\ end {align}

Through further summaries, the change in the Gibbs energy of a (wetting) nucleus assumed to be spherical with the radius \ (r \) for the given undercooling \ (\ Delta T \) can be represented as follows:

\ begin {align}

\ label {heterogeneous}

\ boxed {\ Delta G_ {Keim} (\ Delta T, r) = \ tfrac {1} {4} \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) ~ \ left [- \ frac {4 ~ \ pi ~ \ rho ~ q_E} {3 ~ T_E} ~ \ Delta T ~ r ^ 3 + 4 \ pi ~ \ gamma_ {KS} ~ r ^ 2 \ right]} \ [5px]

\ end {align}

This equation is discussed in more detail in the following section.

### Comparison of the heterogeneous with the homogeneous nucleation

A comparison with the change in energy of a comparable nucleus during homogeneous nucleation is interesting at this point:

\ begin {align}

\ label {germ}

& \ boxed {\ Delta G_ {hom} (\ Delta T, r) = - \ frac {4 \ pi \ rho ~ q_E} {3 ~ T_E} \ cdot \ Delta T \ cdot r ^ 3 + 4 \ pi ~ \ gamma \ cdot r ^ 2} \ [5px]

\ end {align}

With the same radius and the same undercooling, the change in energy in the case of heterogeneous nucleation differs by the geometric factor in the square brackets of the equation (\ ref {heterogeneous}) in front of the term (wetting factor).

The term in the square brackets corresponds to the change in energy in the case of homogeneous nucleation. From an energetic point of view, the heterogeneous nucleation on a smooth wall is linked to the homogeneous nucleation as follows:

\ begin {align}

\ boxed {\ Delta G_ {het} = f (\ Theta) \ cdot \ Delta G_ {hom}} ~~~ \ text {with} ~~~ \ boxed {f (\ Theta) = \ tfrac {1} { 4} \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right)} \ le 1 \ [5px]

\ end {align}

The critical nucleus radius in heterogeneous nucleation is basically the same as in homogeneous nucleation, only the activation energies differ (under otherwise identical conditions). Note that the heterogeneous nucleation involves the same (fictitious) nucleus radius, but ultimately only part of the spherical volume actually forms the nucleus.

The critical nucleus radius can in principle be obtained in the same way as for homogeneous nucleation by deriving and setting the equation (\ ref {heterogeneous}) to zero:

\ begin {align}

& \ frac {d (\ Delta G _ {\ text {Keim}})} {dr} = 0 \ [5px]

& \ tfrac {1} {4} \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) \ cdot \ left [- \ frac {4 \ pi \ rho ~ q_E} {T_E} \ cdot \ Delta T \ times r ^ 2 + 8 \ pi \ gamma_ {KS} \ times r \ right] = 0 \ [5px]

& \ frac {4 \ pi \ rho ~ q_E} {T_E} \ cdot \ Delta T \ cdot r = 8 \ pi \ gamma_ {KS} \ [5px]

& \ boxed {r_ {k, het} = \ frac {2 \ gamma_ {KS} ~ T_E} {q_E ~ \ rho ~ \ Delta T}} = r_ {k, hom}

\ end {align}

Inserting the critical nucleus radius \ (r_ {k, het} \) into equation (\ ref {heterogeneous}) provides the activation energy of heterogeneous nucleation \ (\ Delta G_ {k, het} \), with the already explained relationship to homogeneous nucleation :

\ begin {align}

& \ boxed {\ Delta G_ {k, het} = \ tfrac {1} {4} \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right) \ cdot \ left [\ frac {16 ~ \ pi ~ \ gamma_ {KS} ^ 3 ~ T_E ^ 2} {3 ~ q_E ^ 2 ~ \ rho ^ 2 ~ \ Delta T ^ 2} \ right]} \ [5px]

\ end {align}

or.

\ begin {align}

& \ boxed {\ Delta G_ {k, het} = f (\ Theta) \ cdot \ Delta G_ {k, hom}} ~~~ \ text {with} ~~~ \ boxed {f (\ Theta) = \ tfrac {1} {4} \ left (2-3 \ cos \ Theta + \ cos ^ 3 \ Theta \ right)} \ le 1 \ [5px]

\ end {align}

The term that depends purely on the wetting angle is always less than 1, so that it can be seen at this point that the heterogeneous nucleation requires a lower activation energy than the homogeneous nucleation.

For a wetting angle of \ (\ Theta \) = 90 ° the nucleus is a hemisphere and the wetting factor has the value \ (f (\ Theta) \) = 0.5. In this case, the nucleus only needs half the activation energy, after all it has only half the volume compared to a spherical nucleus with homogeneous nucleation.

With larger wetting angles, the wetting factor increases more and more. At an angle of \ (\ Theta \) = 180 °, this wetting factor finally reaches its maximum of 1. The spherical segment has become a round sphere that no longer wets the wall. As a special case, homogeneous nucleation is obtained!

Conversely, for small wetting angles, i.e. for very strong wetting, very small wetting factors are obtained. The activation energy required for nucleation therefore drops very sharply. With a smaller wetting angle, the germ volume then also decreases very sharply with the same critical radius. So many atoms no longer have to be made available for nucleation.

At a wetting angle of \ (\ Theta \) = 20 °, for example, the required number of atoms decreases from 300,000 atoms to only 800 atoms compared to the example already presented in the case of homogeneous nucleation to only 800 atoms (under otherwise identical conditions).

The activation energy increases by the same factor from the original 9.5 · 10^{-16} J to 0.025 x 10^{-16} J from. This makes it clear once again that heterogeneous nucleation is much more likely than homogeneous.

The heterogeneous nucleation works particularly well when the introduced foreign bodies (seed particles) or the vessel walls can be wetted very well.

When introducing seed crystals into the melt, care must be taken that the seed particles have a larger radius than that of the critical radius of nucleation. At the same time, however, there is a risk that the particles will dissolve in the hot melt before nucleation could have started. For this reason, the seed particles are often introduced into the supercooled melt as soon as it is poured.

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