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This animation shows several applications of Fick's 2nd law. Select a scenario and click "Continue" to proceed.

Interstitial
Impurity

Metal

In this case of a finite source for diffusion, we have an interstitial impurity located at the end of our piece of metal. We assume that it has a negligible width compared to the width of the metal.

In this situation the interstitial impurity will diffuse through the metal and the metal will not diffuse at all

Fick's 2nd law
$\frac{{\partial C}}{{\partial t}} = D\frac{{{\partial ^2}C}}{{\partial {x^2}}}$
In this case the boundary conditions for solving Fick's 2nd law are:

$$C(x > 0,t = 0) = 0$$ (At the start, there are no impurity atoms in the metal)

$$\int {C(x,t).dx = B}$$  (The total amount of impurity atoms remains constant)

Where C(x,t) is the concentration of the impurity. The solution for C is:

$C(x,t) = \frac{B}{{\sqrt {\pi Dt} }}\exp \left( { - \frac{{{x^2}}}{{4Dt}}} \right)$

The value D is the
diffussivity of the atoms
$C(x,t) = \frac{B}{{\sqrt {\pi Dt} }}\exp \left( { - \frac{{{x^2}}}{{4Dt}}} \right)$

D

t =

In this case we have H2 diffusing into steel from one side. We assume that the steel bar has a semi infinite length and that there will be a constant concentration of H2 at the interface of the steel and the H2. From now on we will represent the diffusion with a schematic diagram (top right picture)

In this situation the hydrogen will diffuse into the steel and the steel will not diffuse at all

Fick's 2nd law
$\frac{{\partial C}}{{\partial t}} = D\frac{{{\partial ^2}C}}{{\partial {x^2}}}$

In this case the boundary conditions for solving Fick's 2nd law are:

$$C(x > 0,t = 0) = 0$$ (At the start, there are no impurity atoms in the metal)

$$\int {C(x = 0,t) = B}$$ (The concentration at x=0 is constant)

Where C(x,t) is the concentration of the impurity. The solution for C is:

$C(x,t) = B - {\mathop{\rm e}\nolimits} rf\left( {\frac{x}{{2\sqrt {Dt} }}} \right)$

The value D is the diffussivity of the atoms
erf is the error function
$C(x,t) = B - {\mathop{\rm e}\nolimits} rf\left( {\frac{x}{{2\sqrt {Dt} }}} \right)$

D

t =

In this case we have two semi-infinite bars placed beside each other. We assume that both metals will undergo substitutional diffusion and will therefore diffuse into each other. We also assume that the two metals have the same diffusivity.
From now on we will using the schematic diagram of the two metals (top right picture)

Fick's 2nd law
$\frac{{\partial C}}{{\partial t}} = D\frac{{{\partial ^2}C}}{{\partial {x^2}}}$
The solution for this situation is similar to the solution for a constant source. One main difference is that at the interface the concentration is constant at B (50% Metal 1, 50% Metal 2). This is because the diffusivity of both the metals is equal so the flux of both metals across the boundary is equal. In addition we say that the concentration of Metal 1 is 100% at -∞ and the concentration of Metal 2 is 100% at +∞
In our solution we consider the concentration of Metal 1, C1, which will have a maximum concentration of 100.
The concentration of Metal 2 will then be C2(x,t) = 100 - C1(x,t)
The solution is
$C(x,t) = \frac{B}{{\sqrt {\pi Dt} }}\exp \left( { - \frac{{{x^2}}}{{4Dt}}} \right)$

The value D is the diffussivity of the atoms
$C(x,t) = \frac{B}{{\sqrt {\pi Dt} }}\exp \left( { - \frac{{{x^2}}}{{4Dt}}} \right)$
Metal 1     Metal 2

D

t =