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

Answer 1

Answer: (A)

$\frac{Ze}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{r^{2}} - \frac{r}{R^{3}} \right\rbrack$

Answer 2

: Charge on nucleus is, = + Ze

Total negative charge = - Ze

($\because$atoms is electrical neutral

Negative charge density, $\rho = \frac{ch\arg e}{volume}$

$\frac{- Ze}{\frac{4}{3}\pi R^{3}}$

i.e. $\rho = - \frac{3}{4}\frac{Ze}{4\pi R^{3}}$ …(i)

Consider a Gaussian surface with radius r. By Gauss’s theorem

$\varphi = E(r) \times 4\pi r^{2} = \frac{q}{\varepsilon_{0}}$ …(ii)

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)

$E(r) = \frac{q'}{4\pi\varepsilon_{0}r^{2}} = \frac{Ze - Ze\frac{r^{3}}{R^{3}}}{4\pi\varepsilon_{0}r^{2}} = \frac{Ze}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{r^{2}} - \frac{r}{R^{3}} \right\rbrack$