The binding energy is related to the kinetic energy by the equation BE = hv - KE, where hv is your photon energy (typically 1486.6 eV for Al ka radiation) and KE is the meansured emmited electrons kinetic energy

Most software tends to allow you to swap easily between these scales so calculation isnt required

As for teh energy step, I am confused - if you mean the step szie between data points, then a larger step will give you less definition and therefore less accuracy.   Usually wide scans are perfromed at a step of 1 eV at a high pass energy (therefore low resolution) where as high resolution spectra may be done at say 20 eV pass energy with a 0.1 eV step, therfore giving you much more accuracy.

thanks.

Yes i mean the step size between data points.  In our case it is 0.8 eV.

Then that is no good for accurate binding energy determination. Should be 0.1 at least for high resolution scans.

I suspect you are only looking at wide/survey scans with that step size so do not take the binding energy as accurate

Would you please help me for question (d), I know that Eb = 403.7 but don't know how to prove it .

An XPS electron was found to have a kinetic energy of 1055.3 eV when ejected with an Al Kα source (λ= 0.83393 nm) and measured in a spectrometer with a work function of 25.1 eV. The electron is believed to be a N (1s) electron in NaNO3. (a) What was the binding energy for the electron? (b) What would be the kinetic energy of the electron if a Mg Kα (λ=0.98900 nm) source were used? (c) How could one be sure that a peak was an XPS and not an Auger electron peak? (d) At what binding and kinetic energies would a peak for NaNO2 be expected when the Al Kα source was used with the same spectrometer?

Solution :

• h is Planck constant ( 6.62 x 10-34 Jꞏs )

• ν is the frequency (Hz) of the radiation

• c is the speed of light (2.998 x 108 m/s)

1 J = 6.242 x 1018 eV

(a)

E= hʋ= hc/

- Al Kα radiation (0.83393 nm): calculate the energy?

6.62 x 10-34 x 2.998 x 108

E = ____________________________________ = 2.3799072 x 10-16 J

0.83393 10-9

Then, we convert the energy to eV, so:

2.3799072 x 10-16 x 6.242 x 1018 = 1485.53 eV

Eb= hʋ – Ek – w

Eb= 1485.53 - 1055.3 – 25.1 = 405.13 eV

(b)

Mg Kα (λ=0.98900 nm), so:

6.62 x 10-34 x 2.998 x 108

E= ____________________________ = 2.0067503 x 10-16 J

0.98900 10-9

Then, we convert the energy to eV, so:

2.3799072 x 10-16 x 6.242 x 1018 = 1252.61 eV

Ek = 1252.61 – 405.13 – 25.1 = 822.38 eV

(c)

Using different sources with differing energies, such as the Al and Mg ones. Obtained Auger peaks do not change with the two sources, however, XPS peaks change.

(d) ???