= Solution
Place the interface at $x=0$ and let the incident beam travel in the positive x direction. Set the vacuum potential to zero and the solid potential lower by $V_0>0$:
$$
\boxed{V(x)=\begin{cases}0&x<0,\\-V_0&x\geq0.\end{cases}}
$$
Equivalently, $V(x)=-V_0\Theta(x)$. The incident kinetic energy is $E=mv^2/2$. This ideal <potential step> assumes a sharp, planar interface, constant particle mass and no absorption, inelastic processes or lateral structure. It models the prescribed decrease in potential without asserting that every real solid has the same electronic interface potential.
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