= 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 $K_{a,C}=515.0-284.8-4.4=225.8\ \mathrm{eV}$. Thus
$$
E_{p,C}=\frac{225.8}{1.114}=\boxed{202.69\ \mathrm{eV}}.
$$
The 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: \b[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.
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