= Photoelectron peak overlap and resolution
Choose two hypothetical environments A and B of one <chemical element>, with line centers $\mu_A=284\,\mathrm{eV}$ and $\mu_B=286\,\mathrm{eV}$. Model each line by a normalized <Gaussian function> with common <standard deviation> $\sigma$, and stipulate integrated count weights $w_A=3/4$, $w_B=1/4$. On the <binding energy> axis $E$, the normalized count density is
$$
G(E;\mu,\sigma)=\frac{1}{\sqrt{2\pi}\sigma}
\exp\left[-\frac{(E-\mu)^2}{2\sigma^2}\right],
\qquad
I(E)=\frac34G(E;284,\sigma)+\frac14G(E;286,\sigma).
$$
All energies in this expression use <electronvolts>. Each component's integrated <area> equals its count weight, and $\int I(E)\,dE=1$. There is no background, statistical counting fluctuation, or additional emission feature in this stipulated model. These choices deliberately isolate <photoelectron peak overlap and resolution>.
\Image[/photoelectron-chemical-shift-resolution.png]
{title=Two chemical components can blend into one maximum}
{id=image-photoelectron-chemical-shift-resolution}
{description=Original arbitrary normalized data, not a measured spectrum. The chosen <Gaussian mixture distribution> has centers $284$ and $286\,\mathrm{eV}$ and integrated count weights $0.75$ and $0.25$. Changing the common <standard deviation> from $0.25$ to $1.00\,\mathrm{eV}$ changes the number of maxima from two to one, while leaving the centroid at $284.5\,\mathrm{eV}$. These count weights are not <mole fractions>; a chemical abundance inference requires <cross section> and analyzer-response calibration.}
Direct integration, or the <law of total expectation> applied to the component label, gives the centroid
$$
\bar E=w_A\mu_A+w_B\mu_B=284.5\,\mathrm{eV}.
$$
The centroid lies between the chemical environments and is not either component's <binding energy>. Expanding $(E-\bar E)^2$ within each component, the odd cross term integrates to zero. The resulting <variance> is
$$
\operatorname{Var}(E)=\sigma^2+w_Aw_B(\mu_B-\mu_A)^2
=\sigma^2+0.75\,\mathrm{eV}^2.
$$
For $\sigma=0.25\,\mathrm{eV}$ the <variance> is $0.8125\,\mathrm{eV}^2$ and the two maxima are separated. For $\sigma=1.00\,\mathrm{eV}$ it is $1.7500\,\mathrm{eV}^2$ and there is one maximum. Each Gaussian's <full width at half maximum> is $2\sqrt{2\ln2}\,\sigma$, approximately $0.589$ and $2.355\,\mathrm{eV}$ respectively. The chemical separation remains $2\,\mathrm{eV}$ in both cases. The number of visible maxima therefore does not equal the number of chemical components.
One interpretation of this change is additional independent Gaussian instrument blur. The <convolution of probability densities> adds the <variances>, so broadening $\sigma=0.25$ to $1$ requires response <standard deviation> $\sqrt{1-0.25^2}=\sqrt{15}/4\simeq0.968246\,\mathrm{eV}$. A normalized response with zero mean preserves the centroid and integrated counts. Real line shapes and backgrounds can depart from this model, and separating hidden components is an <inverse problem>; a plausible fit does not alone establish uniqueness.
Finally, count weights describe detection. In a simple response model, let $n_i$ denote the number of emitting sites sampled in environment $i$, $s_i$ its effective <photoionization> <cross section>, and $T_i$ the probability that its emitted <Electron> contributes to the selected measured line. Then
$$
A_i=Cn_i s_iT_i,\qquad
w_i=\frac{n_i s_iT_i}{\sum_jn_js_jT_j}.
$$
The common factor $C$ includes exposure. Here $T_i$ includes escape from the sample, analyzer transmission and detection. Count weights equal site fractions only if the relevant response factors are equal or corrected. They are not automatically molecular <mole fractions>; different <molecules> can supply different numbers of the selected emitting sites. In an original example, equal site numbers with $s_A T_A=3s_B T_B$ give the plotted $3:1$ count ratio despite a $1:1$ site ratio.
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