Solution
= Solution
Fixed entrance retardation and fixed pass energy select the same pre-retardation <kinetic energy>, 111.4 eV. The required <photon> energy is therefore
$$
h\nu_C=E_{B,C}+\phi_a+K_a=284.8+4.4+111.4=\boxed{400.6\ \mathrm{eV}}.
$$
It is lower than the original excitation by exactly the difference between the two core <binding energies>, 114.4 eV.