 Solution

The same wave vectors now differ in energy by only 0.070 eV, so the curvature is $A^{\prime }=0.14A=3.64756\mathrm{eV}{\mathring{\mathrm{A}}}^2$. ThereforeThe nearly free-electron in-plane mass and high unoccupied energy suggest an image-potential-derived state, with an electron confined outside the film by its image attraction and surface barrier but relatively free parallel to it. Relative to the supplied clean-platinum vacuum level, its minimum is 0.84 eV below vacuum, a plausible scale for an image state. gold adsorption can change the work function, so 0.84 eV is conditional on using the supplied clean-surface value; an adsorbate work-function measurement is needed for a quantitative assignment. The dispersion alone supports the interpretation without proving it uniquely.