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

Answer 1

The correct option is A $\frac{Ze}{4\pi {ϵ}_0}[\frac{1}{r^2}-\frac{r}{R^3}]$

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 Gauss's 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)
$E\left(r\right)=\frac{q^{\prime }}{4\pi {\epsilon }_0{r}^2}=\frac{Ze-Ze\frac{r^3}{R^3}}{4\pi {\epsilon }_0{r}^2}=\frac{Ze}{4\pi {\epsilon }_0}(\frac{1}{r^2}-\frac{r}{R^3})$