Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section B / 38 / d / Solution

Define the retard ratio using the kinetic energy at the analyser entrance before the deliberate retardation: $R=K_a/E_p$. Initially $R=111.4/100.0=1.114$. With the original excitation, a carbon photoelectron arrives with $K_{a,C}=515.0-284.8-4.4=225.8\mathrm{eV}$. ThusThe new deliberate retardation is $225.8-202.69=23.11\mathrm{eV}$. For fixed slits, angular acceptance and hemisphere radii, the geometrical fractional energy width is approximately constant, so the absolute analyser width grows in proportion to pass energy: about 2.03 times poorer absolute energy resolution. Counts may improve, but closely separated chemical shifts become harder to resolve. An alternative definition $(K_a-E_p)/E_p$ gives 0.114 rather than 1.114 for the same physical setting and leads to the same final pass energy when used consistently.