# Kinetic energy of electrons emitted in Photoelectron Spectroscopy

In multiple sources (example), I found the information that the kinetic energy $E_k$ measured using (Ultraviolet / X-ray) Photoelectron Spectroscopy is given by:

$E_k = h\nu - B.E._F - \phi_{spec}$

where $h\nu$ is the energy carried by the incident photon, $B.E._F$ the energy difference between the energy level of the electron and the Fermi energy and $\phi_{spec}$ the work function of the spectrometer.

As far as I understood the measurement process, the actually measured quantity in Photoelectron Spectroscopy is the kinetic energy $E_k$ of the electron. This energy is measured by the radius of the electron in a homogeneous electric field. Thus, the kinetic energy of the electron after the emission would $E_k = h\nu - B.E._F - \phi_{sample}$, this independent of the work function of the spectrometer. So where does this difference come from?

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$$E_\text{binding} = E_\text{photon} - \left(E_\text{kinetic} + \phi\right)$$
where $E_\text{binding}$ is the binding energy (BE) of the electron, $E_\text{photon}$ is the energy of the X-ray photons being used, $E_\text{kinetic}$ is the kinetic energy of the electron as measured by the instrument and $\phi$ is the work function dependent on both the spectrometer & the material. This equation is essentially a conservation of energy equation. The work function term $\phi$ is an adjustable instrumental correction factor that accounts for the few eV of kinetic energy given up by the photoelectron as it becomes absorbed by the instrument's detector. It is a constant that rarely needs to be adjusted in practice.
So, it is the absorbtion of the photoelectron by the detector of the spectrometer that causes the work function term. The work function of the measured sample should be part of $E_\text{binding}$ as this is the ionisation energy of the electron. However, I think the equation you quote might have it slightly wrong because if $B.E._F$ is the energy difference between the energy level of the electron and the Fermi energy then there is indeed some portion of the work function of the sample missing since $B.E._F$ is not equal to the ionisation energy.