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. $\frac{Ze}{4\pi {\epsilon }_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right]$
• B. $\frac{Ze}{4\pi {\epsilon }_0}\left[\frac{1}{r^3}-\frac{r}{R^2}\right]$
• C. $\frac{Ze}{4\pi {\epsilon }_0}\left[\frac{r}{R^3}-\frac{1}{r^2}\right]$
• D. $\frac{Ze}{4\pi {\epsilon }_0}\left[\frac{r}{R^3}+\frac{1}{r^2}\right]$

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,
$\varphi =E\left(r\right)\times 4\pi r^2=\frac{q}{{\epsilon }_0}...\left(ii\right)$
Charge enclosed by Gaussian surface
$q^{\prime }=Ze+\frac{4\pi r^3}{3}\rho =Ze-Ze\frac{r^3}{R^3}$ (Using (i)
From $(ii)$,
$E\left(r\right)=\frac{q^{\prime }}{4\pi {\epsilon }_0{r}^2}$
$=\frac{Ze-Ze\frac{r^3}{R^3}}{4\pi {\epsilon }_0{r}^2}$
$=\frac{Ze}{4\pi {\epsilon }_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right]$