An atom has a positively charge point nucleus of charge Ze surrounded by a uniform density of negative charge upto a radius R. The atom is taken as neutral. The electric field at a distance r<R from the nucleus is given by-

$\frac{Ze}{4πε_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right]$

The correct answer is Option (2) → $\frac{Ze}{4πε_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right]$

The charge density is related to total charge and the volume of the sphere,

$ρ= \frac{-Ze}{\frac{4}{3} \pi R^3}$

Charge enclosed by the Gaussian surface at distance r,

$Q_{enc}=Ze+ρ.\frac{4}{3} \pi r^3$

Substituting $ρ$ in this equation,

$Q_{enc}=Ze\left(1-\frac{r^3}{R^3}\right)$

According to Gauss's law,

$E.4\pi r^2=\frac{Q_{enc}}{ε_0}$

$E=\frac{Ze}{\frac{4}{3} \pi r^2ε_0}\left(1-\frac{r^3}{R^3}\right)$