Titanium has a hexagonal crystal structure. Consequently, its single crystal diffusion coefficient is orthotropic; that is, D₁₁ = D₂₂ ≠ D₃₃ with the components D₁₁ and D₂₂ lying in the basal plane of the hexagonal lattice and D₃₃ parallel to the c axis. Suppose D₃₃ is half the value of D₁₁ and a oncentration gradient is applied along a direction oriented 45° from the basal plane. At what angle does the resulting flux make with the basal plane?

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To find the angle at which the resulting flux makes with the basal plane, we first need to resolve the concentration gradient vector into components parallel and perpendicular to the basal plane.

Let's denote the concentration gradient vector as G. Since it's oriented 45° from the basal plane, we can represent it as G = G₀(cos(45°), sin(45°), 0), where G₀ is the magnitude of the concentration gradient.

Now, let's denote the flux vector as J. We know that the flux in the basal plane (D₁₁ and D₂₂ directions) will be greater than the flux along the c axis (D₃₃ direction).

So, let's assume the flux vector has components J₁₁, J₂₂, and J₃₃ in the directions of D₁₁, D₂₂, and D₃₃ respectively.

Given that D₃₃ = 0.5*D₁₁, and D₁₁ = D₂₂ ≠ D₃₃, we can write:

J₁₁ = -D₁₁ * ∂C/∂x

J₂₂ = -D₂₂ * ∂C/∂y

J₃₃ = -D₃₃ * ∂C/∂z

Since the concentration gradient is along the 45° direction, we have:

∂C/∂x = G₀cos(45°)

∂C/∂y = G₀sin(45°)

Now, let's compute the flux components:

J₁₁ = -D₁₁ * G₀cos(45°)

J₂₂ = -D₂₂ * G₀sin(45°)

J₃₃ = -D₃₃ * 0 (because the gradient is along the basal plane)

To find the angle that the resulting flux makes with the basal plane, we can take the dot product of the flux vector with a vector in the basal plane and find the angle between them.

The flux vector in the basal plane is (J₁₁, J₂₂, 0), and a vector in the basal plane is (1, 1, 0) for convenience.

Let's find the dot product:

(J₁₁, J₂₂, 0) dot (1, 1, 0) = J₁₁1 + J₂₂1 + 0*0

= J₁₁ + J₂₂

Substitute the expressions for J₁₁ and J₂₂:

= -D₁₁ * G₀cos(45°) - D₂₂ * G₀sin(45°)

Now, since D₁₁ = D₂₂, we can factor that out:

= -(D₁₁ + D₂₂) * G₀*cos(45°)

= -2 * D₁₁ * G₀*cos(45°)

Now, the dot product of two vectors is given by the product of their magnitudes and the cosine of the angle between them. So:

J dot basal plane vector = |J| * |basal plane vector| * cos(θ)

Where |J| is the magnitude of J, |basal plane vector| is 1 (since it's a unit vector), and θ is the angle between J and the basal plane.

Set these equal:

-2 * D₁₁ * G₀*cos(45°) = |J| * 1 * cos(θ)

Now, solve for θ:

cos(θ) = -2 * D₁₁ * G₀*cos(45°) / |J|

θ = arccos(-2 * D₁₁ * G₀*cos(45°) / |J|)

Substitute the expression for |J|:

θ = arccos(-2 * D₁₁ * G₀cos(45°) / sqrt((D₁₁G₀cos(45°))^2 + (D₂₂G₀*sin(45°))^2))

Now, plug in the given values for D₁₁, D₂₂, and G₀ to find the angle theta