The spin-orbit interaction energy ($H_{SO}$) for an electron in a central field is proportional to the dot product of orbital and spin angular momenta ($\vec{l}\cdot\vec{s}$) and depends on the radial part $f(r)$ as given: $H_{SO} = f(r)\vec{l}\cdot\vec{s}$.
The radial function $f(r)$ is determined by the potential $V(r)$ created by the charge distribution. The standard form is:
$f(r) \propto \frac{1}{r}\frac{dV(r)}{dr}$Consider an electron moving inside a uniformly charged sphere of radius $R$. The electric field $E(r)$ at a distance $r$ from the center ($r < R$) depends on the enclosed charge. Using Gauss's Law, the electric field is found to be proportional to the radius:
$E(r) \propto r$The electric field is related to the potential $V(r)$ by $E(r) = -\frac{dV(r)}{dr}$. Therefore, the gradient of the potential has the following radial dependence:
$\frac{dV(r)}{dr} = -E(r) \propto -r$Now, substitute the radial dependence of the potential gradient into the expression for $f(r)$:
$f(r) \propto \frac{1}{r}\frac{dV(r)}{dr}$ $f(r) \propto \frac{1}{r}(-r)$ $f(r) \propto -1$The result $f(r) \propto -1$ shows that $f(r)$ is a constant value, independent of the radial distance $r$. This corresponds to Option A.