Question

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 $(r < R)$

• A. $\frac{Ze}{4\pi\varepsilon_{0}}\left[\frac{1}{r^{2}}-\frac{r}{R^{3}}\right]$
• B. $\frac{Ze}{4\pi\varepsilon_{0}}\left[\frac{1}{r^{3}}-\frac{r}{R^{2}}\right]$
• C. $\frac{Ze}{4\pi\varepsilon_{0}}\left[\frac{r}{R^{3}}-\frac{1}{r^{2}}\right]$
• D. $\frac{Ze}{4\pi\varepsilon_{0}}\left[\frac{r}{R^{3}}+\frac{1}{r^{2}}\right]$

Answer 1

Answer: (A)

Charge on nucleus is $= + Ze$
Total negative charge$ = - Ze$ ($\because$ atoms is electrical neutral)
Negative charge density, $\rho=\frac{charge}{volume}=\frac{-Ze}{\frac{4}{3}\pi R^{3}}$
i.e., $\rho=-\frac{3}{4} \frac{Ze}{\pi R^{3}}\,...\left(i\right)$
Consider a Gaussian surface with radius $r$.
By Gausss theorem,
$\phi=E\left(r\right)\times4\pi r^{2}=\frac{q}{\varepsilon_{0}}\,...\left(ii\right)$
Charge enclosed by Gaussian surface
$q'=Ze+\frac{4\pi r^{3}}{3} \rho=Ze-Ze \frac{r^{3}}{R^{3}}$ (Using (i)
From $(ii)$,
$E\left(r\right)=\frac{q'}{4\pi\varepsilon_{0}r^{2}}$
$=\frac{Ze-Ze \frac{r^{3}}{R^{3}}}{4\pi\varepsilon_{0}r^{2}}$
$=\frac{Ze}{4\pi\varepsilon_{0}}\left[\frac{1}{r^{2}}-\frac{r}{R^{3}}\right]$