Electric Field at Nucleus Boundary for Non-Uniform Charge Distribution

To find the electric field at the boundary of the nucleus ($r = R$), we apply Gauss's Law to a spherical Gaussian surface of radius $R$ centered at the origin of the nucleus.

According to Gauss's Law, the total electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the permittivity of free space ($\varepsilon_0$):

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

Since the charge density $\rho(r)$ is spherically symmetric, the electric field $\vec{E}$ at $r = R$ is purely radial and has a uniform magnitude $E(R)$ over the entire surface of the sphere of radius $R$.

The area of the spherical surface at $r = R$ is $4\pi R^2$. Therefore, the left-hand side of Gauss's Law simplifies to:

$\oint \vec{E} \cdot d\vec{A} = E(R) \cdot 4\pi R^2$

The total nuclear charge enclosed within the nucleus of radius $R$ is given to be $Ze$, which is a constant:

$Q_{\text{enclosed}} = Ze$

Substituting this into Gauss's Law yields:

$E(R) \cdot 4\pi R^2 = \frac{Ze}{\varepsilon_0}$

Solving for $E(R)$:

$E(R) = \frac{Ze}{4\pi \varepsilon_0 R^2}$

From this equation, it is evident that the electric field at $r = R$ depends only on the total nuclear charge $Ze$ and the overall radius $R$. It is completely independent of the internal distribution parameter $a$.

Therefore, the electric field at $r = R$ is independent of $a$.

Correct Option: A