 Solution

A step elastically reflects a surface-state wave. Interference between $e^{ikx}$ and $r{e}^{-ikx}$ produces a local density oscillation at $2k$, so the measured spatial period is ${\lambda }_{\mathrm{osc}}=\pi /k$, not $2\pi /k$. Thus $k_1=\pi /20=0.15708{\mathring{\mathrm{A}}}^{-1}$ and $k_2=\pi /15=0.20944{\mathring{\mathrm{A}}}^{-1}$. Write $E=E_0+A{k}^2$ with $A={\hslash }^2/(2m^{\ast })$. The two points giveHere ${\hslash }^2/(2m_e)=3.80998\mathrm{eV}{\mathring{\mathrm{A}}}^2$. The fitted minimum is $E_0=1.00-A(\pi /20{)}^2=\boxed{0.35714\mathrm{eV}}$ above the Fermi level. This assumes the bias measures sample excitation energy, parabolic dispersion over the measured interval, and no wave-vector-dependent phase contribution that changes the fitted period.