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Checking the math in RAW's mas 88 paper

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Treating each electron individually and neglecting absorption effects we can estimate the thickness from eq 8.

We must also be assuming an element is only contained in 1 layer:

(1)ki,exp=0zif1ϕ(ρz)spd(ρz)0ϕ(ρz)std(ρz)

But we don’t know the phi(rho z) until we iterate so we use a triangular approximation:

phirhoz-A1

The area of the shaded region = qρzif1[(ρzif1)2q/ρzr,i]/2

The area of the big triangle = [qρzr,i]/2

For simplicity lets call:

ρzr,i=R
ρzif1=x

So we have:

k=qxx2qR12qR12
(2)k=2Rx2R2
x=2R±4R24kR22R2
(3)x=R±R1k

So Waldo’s final answer was correct… for a surface film.

But What about for a buried layer?

The actual (\phi(\rho z)) curve whould be shifted but the idea is the same:

phirhoz-A2

So:

For simplicity let’s call:

ρzr,i=R
ρzif1=x
ρzif2=y

The area of the blue box (second region) is:

(qxqR)(yx)(yx)2qR12

So we have:

k=(qxqR)(yx)(yx)2qR12qR12k=2y2xR2xyR2+2x2R2y2R2+2xyR2x2R2
(4)k=2yRy2R22xR+x2R2
y=(2R)±(2R)24R2(k+x2R22xR)2R2
(5)y=R±R1k2xR+x2R2

Or:

(6)δ0fn=iRi(1.01ki,exp2xR+(x)2R2)x So GMRFilm is correct but if you look the paper is wrong :(

Finding starting compositions

For surface film and burried layer we can use eqs. 2 and 4 respectivly.

For the substrate we substitute R for y in eq. 4.

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