Question

The radius of a gold nucleus (Z = 79) is about $7·0\times {10}^{-10}\mathrm{m}$. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?

Answer 1

Given:
Atomic number of gold = 79
Charge on the gold nucleus, $Q=79\times (1.6\times {10}^{-19})\mathrm{C}$

The charge is distributed across the entire volume. So, using Gauss's Law, we get:

(a)

$$\begin{gathered} \varphi =\oint \overrightarrow{\mathrm{E}}.d\overrightarrow{s}=\frac{Q}{{\in }_0} \\ \Rightarrow \oint Eds=\frac{Q}{{\in }_0} \end{gathered}$$
The value of E is fixed for a particular radius.
$$\begin{gathered} \Rightarrow E\oint ds=\frac{Q}{{\in }_0} \\ \Rightarrow E\times 4\pi r^2=\frac{Q}{{\in }_0} \\ \Rightarrow E=\frac{Q}{{\in }_0\times 4\pi r^2} \end{gathered}$$
$$\begin{gathered} E=\frac{79\times (1.6\times {10}^{-19})}{(8.85\times {10}^{-12})\times 4\times 3.14\times (7\times {10}^{-10}{)}^2} \\ E=2.315131\times {10}^{21}\mathrm{N}/\mathrm{C} \end{gathered}$$

(b) To find the electric field at the middle point of the radius:

Radius, $r=\frac{7}{2}\times {10}^{-10}\mathrm{m}$
Volume,
$$\begin{gathered} V=\frac{4}{3}\mathrm{\pi }{r}^3 \\ =\frac{4}{3}\times \frac{22}{7}\times \frac{343}{8}\times {10}^{-30} \end{gathered}$$

Net charge $=79\times 1.6\times {10}^{-19}\mathrm{C}$

Volume charge density

$=\frac{79\times 1.6\times {10}^{-19}}{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 343\times {10}^{-30}}$

So, the charge enclosed by this imaginary sphere of radius r =3.5 $\times {10}^{-10}$
$$\begin{gathered} =\frac{79\times 1.6\times {10}^{-19}}{{\displaystyle \frac{4}{3}}\mathrm{\pi }\times 343\times {10}^{-30}}\times \frac{4}{3}\mathrm{\pi }\times \frac{343}{8}\times {10}^{-30} \\ =\frac{79\times 1.6\times {10}^{-19}}{8} \end{gathered}$$
$$\begin{gathered} \Rightarrow E=\frac{79\times 1.6\times {10}^{-19}}{8\times 4\mathrm{\pi }{\in }_0.{\mathrm{r}}^2}\mathrm{at}r\mathit{}=3.5\times {10}^{-10} \\ =1.16\times {10}^{21}\mathrm{N}/\mathrm{C}. \end{gathered}$$


As electric charge is given to a conductor, it gets distributed on its surface. But nucleons are bound by the strong force inside the nucleus. Thus, the nuclear charge does not come out and reside on the surface of the conductor. Thus, the charge can be assumed to be uniformly distributed in the entire volume of the nucleus.