It is very handly favorite page: XPS-Teach-Yourself. pdf

Thank you a lot of Dear Steven Van Petegem . For XRD as example I tried to do as your advice, I selected angle in Scan window, the program automatically shift Energy window to (ev), my XRD data obtained from machine has 40 kv, when I checked google to convert(https://www.convertunits.com/from/kV/to/EV) kv to ev I found 40 kv equal 4.0e-14 which is out of the range of calculator(30-3000 ev). I think some thing is wrong, what did you do about that?

Best wishes

Thank you Dear Gedvidas Bikulcius did you mean "XPS-Teach-Yourself. pdf" name of software I tried to find it in google but I did not get it.

Thank in advance

Thank you again Steven Van Petegem this our XRD details "(XRD, X’Pert Pro MPD) using Cu Kα radiation (40 kV and 40 mA)"

The other thing is attenuation length need littlie correct to be penetration depth or its totally different.

Thank in advance

Dear Siddig Abuelgasim ,

let me arrange some terms to be not confused here:

You know that the decrease of x-ray intensity I is accoding to the Lambert-Beer law an exponential function of the depth d and the attenuation coefficient µ:

I = Io*exp (-µ*d) with Io the primary intensity, d the depth and µ the linear x-ray attenuation coeffcient, the inverse is named the attenuation length aL.

attenuation length aL= 1/µ.

From this equation one is able to calculate the mean penetration depth dmean of the x-rays in matter.

For an x-ray beam going perpendicular into the matter in one direction only

it turns out that dmean = 1/µ.

At this depth d= dmean the intensity has dropped down to I = Io*1/e ~37% of Io.

So here 'penetration depth' and 'attenuation length' have the same value of 1/µ.

For the XRD case you do not have the case of a one direction propagation.

The diffracted x-ray beam is scattered (reflected) at the lattice planes and thus changes the direction, is going back to the sample surface and is picked up by the detector.

As it is given in the paper above by Steven Van Petegem , the intensity dependence of the detector can be described by:

I ~ exp(-2µd/sin(theta))

Here (in the XRD case) the mean penetration depth pd is that value of depth d, at which the exponent of the exp function is equal to -1: -2µ*pd/sin(theta)=-1

So we have pd = 1(/2µ) *sin(theta).

The '2' pops up here due to the fact, that the x-ray beam passes twice the sample layer of thickness d and the angle theta shows up due to the non perpendicular geometry.

So please take the µ or 1/µ from your optical constants link and apply

pd = (1/2µ) *sin(theta) for your XRD penetration depth.

You may use also the NIST/XCOM data base for calculation of µ via the x-ray mass-attenuation coefficient µ/rho:

https://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html

Good luck and

best regards

G.M.

Dear prof Gerhard Martens Thank you for this informative comment, as a young researcher I get benefit form each line of your comment.

Regarding sin (theta), is theta here mean the angle at specific peak, i.e., the penetration depth vary along x-axis or this other thing ?

Thank in advance

Dear Siddig Abuelgasim ,

thanks for feedback.

You are right; theta is the angle between the incoming primary x-ray beam impinging onto the XRD sample. The angle between the sample and the diffracted x-ray beam leaving the sample and entering the detector also has the value of theta. So the beam geometry of the XRD set up is just a symmetric reflection set up.

I am sure, you are aware of that issue.

This theta is also the parameter of the x-axis of the XRD pattern, but unfortunately displayed as x=2*theta*) due to historical reasons.

You are right; the penetration depth is not constant for the whole XRD pattern, but varies along the x-axis according to the equation given above.

So the x-ray penetration depth for the 'peaks' (at specific 2*theta positions) is also different according to that equation.

*) 2*theta is the total deflection angle or scatter angle of the primary beam

Additional remark:

the penetration depth (or better 'information depth') for EDX, as you have asked in your question above, is more complicated due to two facts:

a) different angles with respect to the sample surface (e.g. thetap for the primary exciting beam and thetad for the detected beam) and

b) different linear attenuation coefficients for the excitation radiation (e.g. named as µex(Eex)) and the x-ray fluorescence radiation (e.g. named as µfl(Efl)). Both attenuation coefficients depend on the photon energies, which are in use, such as the (mean) excitation energy Eex and the fluorescence energy Efl of the detected x-rays.

The EDX intensity I will have the following dependence on the depth d:

I ~ exp[-d*(µex/sin(thetap) +µfl/sin(thetad))]

Setting the exponent to -1 you will get the penetration (mean information) depth pd as:

pd = 1/[ µex/sin(thetap) +µfl/sin(thetad)]

You see, that for µex = µfl = µ and thetap = thetad = theta the penetration depth formula for the XRD case will show up.

Sorry Peter, I havn't seen your contribution while I was writing and editing my answer here.