Part 3 - Reactivity of solids and nonstoichiometry

Question 1

What is the Kröger-Vink equation for Schottky-defect formation in NiO?

Answer 1

nil = v_Ni ‘’ + v_O**

Question 2

What is the Kröger-Vink equation for Anti-Schottky defect formation in NiO?

Answer 2

Ni_Ni^x + O_O^x = Ni_i ** + O_i ‘’

Question 3

What is the Kröger-Vink equation for Cation Frenkel-formation in NiO?

Answer 3

Ni_Ni^x = v_Ni ‘’ + Ni_i **

Question 4

What is the Kröger-Vink equation for Anion Frenkel-formation in NiO?

Answer 4

O_O^x = v_O ** + O_i ‘’

Question 5

What is the Kröger-Vink equation for intrinsic ionisation in NiO?

Answer 5

nil = e’ + h* (2Ni_ni^x = Ni_Ni ‘ + Ni_Ni *)

Question 6

How can we interpret intrinsic ionisation in a redox-prone oxide such as NiO?

Answer 6

The intrinsic ionisation is given by nil = e’ + h*, and can be interpreted as 2 Ni2+ turns into 1 Ni+ and 1 Ni3+. Also known as disproportionation of divalent nickel into mono- and trivalent defects.

Question 7

What are the possibilities upon oxidation of pure stoichiometric NiO?

Answer 7

Upon oxidation we will get either excess oxygen or less nickel, so we get Ni1-δO or NiO1+δ.

Question 8

What are the possibilities upon reduction of pure stoichiometric NiO?

Answer 8

Upon reduction we will get either excess nickel or deficit oxygen, so we get Ni1+δO or NiO1-δ.

Question 9

In a structure with a large closest-packed atom, what could we say about the likelihood of dominating defects?

Answer 9

Large closest-packed atoms form defects less easily than small atoms in holes, so in these cases one can assume that the defects would be caused by the smaller atoms.

Question 10

How would you go about setting up the equilibrium equations fo oxidative and reductive nonstoichiometry for NiO in O2-gas?

Answer 10

We start by looking at the two alternative compensations of oxidative nonstoichiometry (Ni-vacancies and O-interstitials) and then the two alternative compensations for reductive nonstoichiometry (Ni-interstitials and O-vacancies).

Then we simplify this into final redox equations.

Question 11

If Schottky-defects were the dominant intrinsic structural defects, what would be the defect equilibria (for NiO in O2)?

Answer 11

We would then have equations for Schottky-formation:
nil = v_Ni ‘’ + v_O **

Then we need the intrinsic ionisation equation, and the equations for formation of both kinds of vacancies:

nil = e’ + h*

O2(g) = 2v_Ni'' + 4h* + 2O_o^x
2O_o^x = 2v_o ** + 4e' + O2(g)

Lastly we need the electroneutrality condition as the above four equations only have 3 independent ones:

2[v_Ni ‘’] + [e’] = [h*] + 2[v_o**]

Where the brackets signify concentrations (vacancies have double the charge as the electrons and holes, and as such must be counted twice in the electroneutrality condition).

Question 12

What is the point of integer structure?

Answer 12

This is the point in a Brouwer diagram where there is zero nonstoichiometry, that is there are an equal amount of vacancies of both types (in NiO, [v_o **] = [v_Ni ‘’].

Question 13

What is the point of integer valence?

Answer 13

This is the point where the concentration of electronic defects balance each other out, that is [e’] = [h*].

Question 14

How are the the point of integer structure and point of integer valence related in a pure, binary oxide?

Answer 14

They coincide on the po2-scale.

Question 15

For dominant ionic defects (for a structure where Schottky-defects are the dominant intrinsic structural defects) in NiO, how does the electroneutrality condition look like?

Answer 15

In this case, the concentrations of [e’] and [h*] are so small, that the electroneutrality condition simplifies to [v_o**] = [v_Ni’’] = const.

Question 16

What is the slope of the concentration of [e’] and [h*] for dominant ionic defects in NiO in the central region (around point of integer structure)?

Answer 16

In this case, the number of vacancies are equal each other and constant. We can then see from the mass action terms that log [e’] is proportional to -1/4 log po2 and log [h*] is proportional to 1/4 log po2.

Question 17

For dominant electronic defects (for a structure where Schottky-defects are the dominant intrinsic structural defects) in NiO, how does the electroneutrality condition look like?

Answer 17

In this case, the concentrations of vacancies are so small that the electroneutrality condition simplifies to [e’] = [h*] = const.

Question 18

What is the slope of the concentrations of [v_o**] and [v_Ni’’] for dominant electronic defects in NiO in the central region (around point of integer valence)?

Answer 18

In this case, the number of electronic defects are equal to each other and constant. We can then see form the mass action terms that log [v_O**] is proportional to -1/2 log po2 and that log [v_Ni’’] is proportional to 1/2 log po2.

Question 19

In the most oxidised regions of a log c vs. log po2 plot, how are holes compensated?

Answer 19

Holes are compensated by cation vacancies.

Question 20

In the most reduced regions of a log c vs. log po2 plot, how are electrons compensated?

Answer 20

Electrons are compensated by anion vacancies.

Question 21

How does the electroneutrality condition look for NiO in the most oxidised regions of the log c vs. log po2 plot?

Answer 21

In this area, holes are compensated by nickel vacancies, and we get: [h*] = 2[v_Ni’’].

Question 22

How does the electroneutrality condition look for NiO in the most reduced regions of the log c vs. log po2 plot?

Answer 22

In this area, electrons are compensated by oxygen vacancies, and we get: [e’] = 2[v_O**].

Question 23

What are Brouwer diagrams?

Answer 23

Brouwer diagrams are approximations of a log c (of defects) vs. log po2 plots, using the fact that in many areas the slopes are apporximately linear.

Question 24

If we introduce an acceptor dopant in NiO (e.g. Li+ on Ni2+ sites), how does the mass-action equations change for Shcottky defects?

Answer 24

No change in any equation except the electroneutrality condition. In this case it would be:

2[v_Ni’’] + [e’] + [Li_Ni’] = 2[v_O**] + [h*]

Question 25

What is the effect on the Brouwer diagram on introduction of an acceptor dopant?

Answer 25

When introducing an acceptor dopant two things happen:

1) The point of integer valence is moved towards low po2. This is because Ni2+ is now more easily oxidised in order to keep the charges balanced.
2) The point of integer structure is moved towards high po2. This is because oxygen vacancies are more easily formed in order to keep the charges balanced.

Question 26

What is the effect on the Brouwer diagram of NiO on introduction of a donor dopant?

Answer 26

When introducing a donor dopant two things happen:

1) The point of integer valence is moved towards high po2. This is because NI2+ is now more easily reduced in order to keep the charges balanced.
2) The point of integer structure is moved towards low po2. This is because nickel vacancies are more easily formed in order to keep the charges balanced.

Question 27

Upon doping NiO with an acceptor, what defect dominates around the point of integer structure?

Answer 27

Around the point of integer structure, h* dominates.

Question 28

Upon doping NiO with an acceptor, what defect dominates around the point of integer valence?

Answer 28

Around the point of integer valence, v_O** dominates.

Question 29

Upon doping NiO with an acceptor, what is the electroneutrality condition at the point of integer valence?

Answer 29

At the point of integer valence, oxygen vacancies dominate to compensate the acceptor dopant. This gives us the electroneutrality condition:

2[v_O**] = [Li_Ni’]

Which is constant as we have a constant number of Li.

Question 30

Upon doping NiO with an acceptor, what is the slope of concentration of [h*] and [e’] around the point of integer valence?

Answer 30

As the electroneutrality condition is 2[v_O**] = [Li_Ni’] = const, and this also keeps the concentration of nickel vacancies constant, we can see from the mass action equations that log [h*] must be proportional to 1/4 log po2 and [e’] must be proportional to -1/4 log po2.

Question 31

Upon doping NiO with an acceptor, what is the electronneutrality conidtion at the point of integer structure?

Answer 31

At the point of integer structure, holes dominate to compensate the acceptor dopant. This gives us the electroneutrality condition:

[h*] = [Li_Ni’]

Which is constant as we have a constant number of Li.

Question 32

Upon doping NiO with an acceptor, what is the slope of the concentration of [v_O**] and [v_Ni’’] around the point of integer structure?

Answer 32

As the electroneutrality condition is [h*] = [Li_Ni’] = const, and this also keeps [e’] constant, we can see from the mass action equations that log [v_o**] must be proportional to -1/2 log po2 and log [v_Ni’’] must be proportional to 1/2 log po2.

Question 33

How does the equilibrium equations change for NiO (in the case where Schottky defects are the dominant structural defect) upon donor doping (e.g. of Al3+)?

Answer 33

The mass action terms does not change. The electroneutrality condition changes:

2[v_o] + [h] + [Al_Ni] = 2[v_Ni’’] + [e’]

Question 34

What is the effect of temperature on mixing for isovalent substitutions?

Answer 34

Increasing the temperature will increase entropy term TΔS that drives dissolution, and will thus increase solubility.

Question 35

What is the effect of temperature on mixing for donor substitutions in an oxide?

Answer 35

Increasing the temperature will favour loss of oxygen (chemical reduction) because the released gas has large entropy. A donor substitution makes reduction easier because a donor brings excess oxygen into the system (think Al2O3 into NiO-matrix). Increased temperature will thus favour dissolution of the donor by promoting reduction.

Question 36

What is the effect of temperature on mixing for acceptor substitutions in an oxide?

Answer 36

Increasing the temperature will favour loss of oxygen (chemical reduction) because the released gas has large entropy. An acceptor substitution makes reduction harder because an acceptor brings deficit oxygen (think Li2O into NiO-matrix). Oxygen loss at high temperatures will thus disfavour the dissolution.

However, there is stilll the general entropic driving force of dissolution, and the effect of temperature on solid solubility will not be predicted unless we also consider po2.

Question 37

What is the effect of po2 on mixing for donor substitutiosn in an oxide?

Answer 37

A donor oxide brings excess oxygen, implying a tendency for easy loss of O2, and this will be favoured by low po2.

Question 38

What is the effect of po2 on mixing for acceptor substitutions in an oxide?

Answer 38

An acceptor oxide brings deficit oxygen, implying a tendency for easy inclusion of the missing oxygen into the solid. High po2 will promote this process, and increase the solid solubility of an acceptor.

Question 39

What is the combined effect of temperature and po2 on mixing for donor substitutions?

Answer 39

This is unambiguous. High temperature and low po2 will increase solubility.

At point of integer structure, the donor defect is compensated by [e’]. From the dissolution reaction:

Al2O3 = 2e’ + 2Al_Ni* + 2O_O^x + 1/2O2(g)

we get the mass action term Ksd = [Al_Ni*]2 [e’]^2 po2^1/2

We know that [Al_Ni*] = [e’], so that Ksd = x^4 po2^1/2. Thus decreasing po2 will increase solubility.

For point of integer valence, the donor defect is compensated by [v_Ni’’]. In this case the dissolution reaction is:

Al2O3 = v_Ni’’ + 2Al_Ni* + 3O_O^x

This gives us a mass action term independent on po2, so here the partial pressure doesn’t matter.

Question 40

What is Fick’s first law?

Answer 40

J = D(- dc/dy)

Where J is the flux, c is the concentration of the particle and y is the direction we are considering the movement. D is the diffusion constant.

In other words, the flux is proportional to the negative of the concentration gradient.

Only applicable in steady-state diffusion.

Question 41

What is a flux?

Answer 41

A flux is the magnitude of something passing per unit time through a unit cross-sectional area perpendicular to the flux. Also it is the flow per cross-sectional area.

Some examples include:

1) molecular flux, which is number of atoms per unit time per cross-sectional area.
2) molar flux, which is the number of moles per unit time per cross-sectional area.
3) mass flux, which is the number of kilograms per unit time per cross-sectional area.

Question 42

What is a flow?

Answer 42

A flow is the magnitude of something passing per unit time. E.g. volume of water passing through a river.

Question 43

What is the diffusion coefficient?

Answer 43

The diffusion coefficient is a proportionality constant that shows up in Fick’s laws. It is also called the diffusivity. It is characteristic for a particle-matrix pair at a given temperature and pressure, and is thus not constant.

Question 44

What does it mean that diffusion is steady-state?

Answer 44

That the particle flow in equals the particle flow out.

Question 45

How does the concentration profile look like in steady-state diffusion?

Answer 45

It is linear, goes from high concentration (or temperature) of something to low concentration (temperature).

Question 46

What is Fick’s second law?

Answer 46

∂c/∂τ = ∂/∂y (D ∂c/∂y)

or if D is constant and independent of c:

∂c/∂τ = D (∂^2c / ∂y^2)

Here τ is the time, D is the diffusivity, c is the concentration and y is the length in y-direction.

It states that the change of concentration with time is proportional to the curvature of the concentration gradient. In a curved gradient, fluxes at two points are different and will tend to equalise the concentration.

Question 47

What is non-steady state diffusion?

Answer 47

Non-steady state diffusion is diffusion that is time-dependent and thus changes over time.

Question 48

What happens for non-steady state diffusion into an enclosed space?

Answer 48

Eventually the concentration all over the place is equal, and there is no more concentration gradient to drive any more diffusion, and there is no more net diffusion.

Question 49

What happens for non-steady state diffusion where the particles can leave the bulk?

Answer 49

Eventually we reach steady-state diffusion.

Question 50

What are the transport vehicles in crystal structures?

Answer 50

Vacancies and interstitials.

Question 51

How can we model the probability of a jump attempt to be successfull?

Answer 51

Using Boltzmann statistics.

p = exp(-ΔG/kT)

Where ΔG is the Gibbs-energy barrier of passing through the unfavourable electrostatic environments between the initial site and the new site, k is Boltzmann’s constant and T is the temperature.

Question 52

What is the rate of progression in bulk diffusion?

Answer 52

The rate of progression is the number of successful jumps per second in a given progression direction.

It consists of multiplying the probability of a jump being successfull, with the jump frequency, the probability of a jump being in the correct direction, the probability that a given site is available and a correlation factor that describes the probability of the vacancy it came from not being occupied by any other atoms before it has the chance to go back.

Question 53

In expressions of diffusion, what is p_avail? What will it be for self-diffusion via vacancies in a bcc-structure? What about an fcc-structure?

Answer 53

p_avail is the probability of there being an available vacancy to jump to among the nearest neighbours. It will be Nn * [v], where Nn is number of nearest neighbours and [v] is the vacancy concentration.

For bcc: 8 nearest neighbours, so p_avail = 8[v]
For fcc: 12 nearest neighbours, so p_avail = 12[v]

Question 54

In expressions of diffusion, what is p_dir? What will it be for self-diffusion via vacancies in a bcc-structure? What about an fcc-structure?

Answer 54

p_dir is the probability that any given vacancy is in the positive direction of the flux.

For bcc: Four of the eight nearest neighbours are in the positive direction, so p_dir = 4/8 = 1/2.

For fcc: Four of the twelve nearest neighbours are in the positive direction, so p_dir = 4/12 = 1/3

Question 55

In expressions of diffusion, what is f_c?

Answer 55

f_c is the correlation factor that describes the probability of other neighbours not jumping into the vacancy left by the initial atom. This is important as there is likely not a vacancy waiting for our atom in the new site, and the probability of it jumping back is large if the other atoms don’t do it first.

It is estimated (for Nn > 4) to be f_c = 1 - 2/(Nn-1)

Question 56

What is the difference between the rate of progression for metal ions jumping through vacancies, and the rate of progression of the vacancies themselves?

Answer 56

For diffusion of the vacancies, the p_dir and f_c would be the same. However, the p_avail would be Nn(1-[v]) instead of Nn [v].

The ratio of the two progressions would be [v] / (1 - [v])

The ratio is a consequence of the jump balance, saying that as 1 vacancy jumps 1000 times, 1000 atoms jump once.

Question 57

What is the mean square displacement (σ^2) after n jumps of equal length λ during random walk?

Answer 57

σ^2 = nλ^2

Question 58

What is the root mean square displacement from the origin after n jumps of equal length λ during random walk?

Answer 58

σ = λ√n

Question 59

What happens to the diffusion if an external field is applied?

Answer 59

If an external field is applied, the jumps develop a preferred direction. It lowers the Gibbs barrier of a jump in one direction.

Question 60

What kind of external fields can drive diffusion in a solid?

Answer 60

It could be a chemical potential gradient or an electric-potential gradient.

Question 61

How does the Gibbs barrier of a jump change with an external field?

Answer 61

If an external field is applied, the Gibbs barrier for migration will be lowered by an energy E equal to 1/2λ F, where F is the external force. In the other direction, the barrier is heightened by the same amount.

Question 62

How can you attain the net rate of progression under an external field?

Answer 62

We calculate the rate of progression in the positive direction (where the field lowers the barrier of migration), and subtract the rate of progression in the negative direction.

Question 63

How does one attain the net flux of particles under an external field?

Answer 63

We multiply the net rate of progression (here r_prog * (λ/kT)F, attained from Taylor expansion of the added exponential term that comes from the barrier modifcation) with the length of each jump, and the local volume concentration of atoms:

J = r_prog * (λ^2 c) / (kT) F

This tells us that the flux is dependent on the probabilitty of random progression and that the flux is linearly proportional to the external force (so long as the Taylor approximation holds).

Question 64

What is the driving force of migration in a concentration gradient?

Answer 64

The driving force in a concentration gradient is the gradient of the partial molar Gibbs energy of the atom (the chemical potential):

F = -dμ/dy

μ = μ0 + kT lna

Question 65

How is Fick’s first law related to the progression rate?

Answer 65

We add the driving force in a concentration gradient, F = -dμ/dy, where μ = μ0 + kT lna (and substituting c for a for dilute concentrations of defects).

We then substitute dμ/dy for dμ/dc * dc/dy, and d/dc (μ0 + kT ln c) = kT/c. We then get

F = 1/c * kT (-dc/dy)

We add this to the expression of the flux under an external force, J = r_prog * (λ^2 c) / (kT) F to get:

J = r_prog * (λ^2 c) / (kT) * 1/c * kT (-dc/dy)

J = r_prog λ^2 (-dc/dy)

This is Fick’s first law, with D = r_prog λ^2

Question 66

How can the diffusion coefficient be expressed in terms of the rate of progression and jump length?

Answer 66

D = r_prog λ^2

This can be found from expressing the flux under a concentration gradient.

Question 67

What is the driving force of migration in an electric field?

Answer 67

F = qE, where E is the field strength and q is the charge. E is also expressed as E = -dV/dy, so that:

F = q(-dV/dy)

Question 68

How can we express the electrical conductivity due to a defect i in terms of the driving force under an electric field?

Answer 68

The driving force in an electric field is F = q(-dV/dy). We substitute this into the expression for the flux under an external force (J = r_prog * (λ^2 c) / (kT) F) to get:

J = r_prog * (λ^2 c) / (kT) q (-dV/dy).

We then multiply with the charge per particle to obtain the charge flux (j = qJ):

j = r_prog * (λ^2 c) / (kT) q^2 (-dV/dy) = σ (-dV/dy)

Where σ is the electrical conductivity due to the defect.

Question 69

What is the transference number of the point defect i?

Answer 69

The transference number is the fraction of the total electrical conductivity σ in a solid due to defects.

Question 70

How can one derive Ohm’s law from migration of charged particles under an electric field?

Answer 70

The driving force in an electric field is F = q(-dV/dy). We substitute this into the expression for the flux under an external force (J = r_prog * (λ^2 c) / (kT) F) to get:

J = r_prog * (λ^2 c) / (kT) q (-dV/dy).

We then multiply with the charge per particle to obtain the charge flux (j = qJ):

j = r_prog * (λ^2 c) / (kT) q^2 (-dV/dy) = σ (-dV/dy)

Where σ =r_prog * (λ^2 c) / (kT) q^2 is the electrical conductivity due to the defect.

If the field is constant, (-dV/dy) changes to E and we get:

j = σE, which is Ohm’s law.

Question 71

What is the Nernst-Einstein relation?

Answer 71

The Nernst-Einstein relation is the relation between σ and D when D is expressed as D = r_prog * λ^2 and substituted into the expression for the electrical conductivity:

σ =r_prog * (λ^2 c) / (kT) q^2 => D/kT = σ/cq^2

Question 72

How can the flux J of a defect be expressed using experimentally available parameters?

Answer 72

1) Random diffusivity D: By using the expression for the flux, and substituting in D = r_prog λ^2 to yield

J = Dc / kT * F

2) Electrical conductivity σ: Substituting the Nernst-Einstein relation (D/kT = σ/cq^2) into the equation avoe. We then get:

J = σ/q^2 * F

Question 73

What is the mobility μ of a species i?

Answer 73

The mobility is the speed of a species i per unit intensity of the field that moves it.

Question 74

How can one obtain the mobility μ of species i?

Answer 74

The mobility is the speed of a species i per unit intensity of the field that moves it, so we can obtain it by multiplying each side of the Nernst-Einstein relation by the charge of the species:

Dq/kT = σ/cq

Both sides now have units m/s per V/m. Left-hand side describes it in terms of random diffusivity, while the right-hand describes it in terms of the electrical conductivity.

Question 75

How can one obtain experimentally the electrical conducitvity due to defect i?

Answer 75

You measure the conductivity with selective electrodes that only conducts via the desired defect. E.g. ZrO2 for oxygen, as it is an electrical isolator but an ionic (O2-) conductor.

Question 76

How can one obtain experimentally the random diffusivity of defect i?

Answer 76

You can use tracer-diffusion techniques with isotopes. The isotope is tracable with MS after sectioning the sample or by measuring radioactivity along the diffusion path if a radioactive isotope is used.

The measured diffusivity will be lower by a factor f_c.

Question 77

What is ambipolar diffusion?

Answer 77

Ambipolar diffusion is diffusion of charged defects (e.g. of O2-) that is followed by charged defects of the other type (e.g. of holes, h+).

That means that there is an influx of both negatively charged particles and postively charged particles, together representing the flow of netural atoms.

Question 78

What are the temperature dependence on “running cost” and “manufacture cost” of defects?

Answer 78

For running costs: this is related to the Boltzmann probability of a successful jump, pb = exp(-ΔG/kT). The higher the temperature, the more likely it is that the jump is successful, and the diffusion rate increases.

For making costs: we know that the equilibrium number of defects increases with increasing value of T, according to ΔG = n(ΔHf - TΔSvib) - kT ln Ω. When differentiating this and solving for dΔG/dn = 0, we find that the fractional concentration of vacancies formed is exp(ΔS_vib /k) exp (-ΔHf / kT).

Question 79

How can the temperature dependence of diffusivity of defects be expressed by the Arrhenius equation?

Answer 79

For running costs: this is related to the Boltzmann probability of a successful jump, pb = exp(-ΔG/kT). The higher the temperature, the more likely it is that the jump is successful, and the diffusion rate increases.

For making costs: we know that the equilibrium number of defects increases with increasing value of T, according to ΔG = n(ΔHf - TΔSvib) - kT ln Ω. When differentiating this and solving for dΔG/dn = 0, we find that the fractional concentration of vacancies formed is exp(ΔS_vib /k) exp (-ΔHf / kT).

The total temperature dependence on diffusivity can then be expressed through:

The total temperature dependence on diffusivity can then be expressed through:

D = λ^2 f_c p_dir p_avail pb v

= λ^2 f_c p_dir Nn[vM] exp(-ΔG/RT) v

= λ^2 f_c p_dir Nn v exp(ΔS_vib /R) exp (-ΔHf / RT) exp(-ΔHm/RT) exp(ΔSm/R)

= λ^2 f_c p_dir Nn v exp(ΔS_vib /R +ΔSm/R) exp( -ΔHf / RT -ΔHm/RT)

= D_0 * exp(Ea / RT)

Question 80

How can one determine the activation barrier (Ea) experimentally?

Answer 80

From the Arrhenius equation, D = D_0 * exp(Ea / RT)

By plotting ln D vs 1 /T we can get Ea from least-squares fitting of the linear slope.

Question 81

When do we not have to think about the making costs when looking at the temperature dependence of diffusivity?

Answer 81

When considering foreign interstitial atoms in low concentrations (e.g. carbon in iron)

Question 82

What are the three straightforward cases when the concentration of vacancies [v_i] is known?

Answer 82

1) When the high temperature [v_i] is frozen at lower temperatures.
2) When the concentration of vacancies are determined by the concentration of aliovalent dopants around the point of integer valence (when all dopants are compensated by vacancies)
3) When it is the dominant defect and it is determined by the equilibrium constant K = exp(ΔS/R - ΔH/RT)*
* ΔG = -RT ln K

Question 83

In what cases does Fick’s 2nd law have analytical solutions?

Answer 83

For time-dependent concentrations across simple shapes such as a plate of infinite thickness, plate of finite thickness a parallelepiped, a cylinder or a sphere.

This is provided D is constant.

Question 84

What is the penetration depths?

Answer 84

The penetration depths is the distance √(Dt). This is the characteristic value indicating the extent of penetration of the diffusant.

It is closed to the diffusion half length (that is the length where 50% concentration change has occured in a given time) for diffusions based on the erf-function.

Question 85

What are some things you must be cautious about when it comes to theoretical diffusion vs. diffusion in real materials?

Answer 85

• In our models, we often assume single-point defects being the transport vehjicles. In reality these are often aggregated defect clusters (e.g. dimers of interstitials or vacancies, or larger more complex clusters).
• We disregard ambipolar diffusion.
• D may be dependent on defect concentration, however we assume it to be constant.
• Sample crystallinity and morphology may be of concern, as transport along extended defects are different (or grain boundaries in polycrystalline materials)
• Impurities (that we do not account for) will increase diffusivity by creating point defects as vehicles for the mass transport.

Question 86

What is the driving force behind coalescing of particles?

Answer 86

Minimisation of surface energy.

Question 87

What is sintering?

Answer 87

Sintering is the process of two smaller particles coalescing into a larger one. The driving force is minimisation of surface energy. The dangling bonds of the surface forms bonds with the neighbouring particle and it merges.