An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral. The electric field at a distance r from the nucleus is (r < R)

$\frac{Ze}{4 {πε}_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

$\frac{Ze}{4 {πε}_{0}} [\frac{1}{r^{3}} - \frac{r}{R^{2}}]$

$\frac{Ze}{4 {πε}_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]$

$\frac{Ze}{4 {πε}_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]$

Solution

$\frac{Ze}{4 {πε}_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

Charge on nucleus = + Ze

Total negative charge = - Ze

( $∵$ atom is electrically neutral)

Negative charge density

ρ = $\frac{charge}{volume}$

= $\frac{- Ze}{\frac{4}{3} {πR}^{3}}$

⇒ ρ = $- \frac{3}{4} \frac{Ze}{π R^{2}}$ ...(i)

Consider a Gaussian surface with radius r.

By Gauss's theorem

E (r) × $4 π r^{2} = \frac{q}{ε_{0}}$ .....(ii)

Charge enclosed by Gaussian surface,

q = Ze + $\frac{4 π r^{3}}{3}$ ρ

= Ze $- Ze \frac{r^{3}}{R^{3}}$ [ using (i) ]

From (ii),

E (r ) = $\frac{q}{4 π ε_{0} r^{2}}$

= $\frac{Ze - Ze \frac{r^{2}}{R^{3}}}{4 {πε}_{0} r^{2}}$

⇒ E (r) = $\frac{Ze}{4 {πε}_{0} r^{2}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

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