= Solution
The same <wave vectors> now differ in energy by only 0.070 eV, so the curvature is $A'=0.14A=3.64756\ \mathrm{eV}\,\mathring{\mathrm A}^2$. Therefore
$$
\boxed{m^{*\prime}=1.0445m_e},\qquad
E_0'=5.00-A'(\pi/20)^2=\boxed{4.910\ \mathrm{eV}}\quad\text{above }E_F.
$$
The nearly free-electron in-plane mass and high unoccupied energy suggest an \b[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.
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