= Solution
TODO: Biology deferred at the user's request: the cancer-treatment account, clinical dose choices, treatment planning and clinical interpretation of the numerical model.
For the specified energy-loss model, multiply $dE/dx=-k/E$ by $2E$ and integrate from the entrance:
$$
\frac{d(E^2)}{dx}=-2k,\qquad
\boxed{E(x)^2=E_0^2-2kx.}
$$
The positive <stopping power> is $S=-dE/dx=k/E$. The model reaches zero <kinetic energy> at
$$
\boxed{R=\frac{E_0^2}{2k}=\frac{200^2}{2(770)}\,\mathrm{cm}=25.97\,\mathrm{cm}.}
$$
At the entrance,
$$
\boxed{S(0)=\frac{770}{200}=3.85\,\mathrm{MeV\,cm^{-1}}.}
$$
Half a centimeter before the formal endpoint, $E^2=2k(0.5\,\mathrm{cm})=770\,\mathrm{MeV^2}$, so
$$
\boxed{E=27.75\,\mathrm{MeV},\qquad S=27.75\,\mathrm{MeV\,cm^{-1}}.}
$$
The signed derivatives are the negatives of the quoted <stopping powers>. This <inverse-energy stopping-power model> deposits <energy> more strongly toward the endpoint. Its divergent endpoint derivative is integrable, giving finite total deposited <energy> $E_0$; a real stopping curve includes finite <energy> spread, range straggling, scattering and a more accurate low-energy stopping law. These numbers are consequences of the supplied ideal equation.
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