If adding two positive numbers dividing it through a positive number gives you a negative number, you're doing something wrong, yes.

By the way, always give an actual citation to you method you are using, especially if it's by a current researcher and not established lecture knowledge yet. Is this the reference you are talking about?

https://www.surfacesciencewestern.com/wp-content/uploads/ass10_biesinger.pdf

I don't see anything in that reference which would turn any component of the ratio calculation negative. Assuming you have assigned A1 as Cu0/1+ and A2 as Cu2+, without any recalculation you have a ratio of 1/8.3. However, what did you assign to your B1 component? B2 is probably the shake-up, although its binding energy is somewhat high, but how did you handle B1? I don't see any reference to such a component in Biesinger's work, either.

Dear Jürgen Weippert

Thank you so much for the answer.

I have used the formula mentioned in the paper and also gone through their Youtube videos "https://www.youtube.com/watch?v=DfsxR4m36g8&list=PLo-L_EaxDgrQeTORCKniVrq_v0M3aY79m&index=7"

In the video, B is taken as B1+B2 (as in the excel sheet) I have also taken the same but in my case A- (A1s/Bs)B is negative.

I have also fit the data assuming there is no Cu(0) or Cu(I) and get the following graph.

Ah, so you have taken the sum of B1 and B2, 43.2, as the B value? Well, this indeed leads to a negative ratio. As mentioned in my previous post, I severely doubt that the B1 component is part of the shakeup since the binding energy would be too low for that.

If I look at your 2p1/2 signal, the middle component looks very different inside there but it seems like the shift is constant. Do peaks of other elements also have such a shifted component? Maybe parts of your sample do not get their electrons back. and you have local differential charging.