An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge -Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. a. Show that the electric field strength inside this atom is

$E_{\text{in}}=\frac{Ze}{4\pi {ϵ}_0}\left(\frac{1}{r^2}-\frac{r}{R^3}\right)$

b. What is E at the surface of the atom? Is this the expected value? Explain.

c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field strength at r = 1 2 R?

$Ze$

We use Gauss' Law to find the electric field inside the sphere.

$\Phi =\oint \overrightarrow{E}·\overrightarrow{da}=\frac{Q}{\epsilon 0}$

Where

$\Phi$is the Electric Flux

E is the Electric Field

d a is the Area

Q is the Total Charge

$\epsilon$0 is the Permittivity

The charge inside the sphere from the electron can be evaluated as:

$Q=-Ze$

The total charge enclosed in a sphere of radius r is $\frac{-Ze{r}^3}{R^3}$

r is the radius inside the atom.

The electric field from the electrons can thus be given by:

$E\left(4\pi r^2\right)=\frac{-Ze{r}^3}{\epsilon 0R^3}$

The electric field from nucleus is from a point charge Z e It is thus being given by:

$E\left(4\pi r^2\right)=\frac{Ze}{\epsilon 0}$

Combining the above equations, we get:

$E=\frac{Ze}{4\pi \epsilon 0}\left(\frac{1}{r^2}-\frac{r}{R^3}\right)$

Since atoms are neutral therefore it can be said that the electric field strength at the surface of the atom is 0

The electric field strength at the uranium surface is calculated as

$E=\frac{Ze}{4\pi \epsilon 0}\left(\frac{1}{r^2}-\frac{r}{R^3}\right)$

Calculation:

Evaluating electric field for the values of

$\mathrm{Z}=92$, $\mathrm{R}=0.1$,$\mathrm{nm}$, r=0.5 R

$E=\frac{92\times 1.6\times {10}^{-19}}{4\pi \times 8.85\times {10}^{-12}}\times \left(\frac{1}{0.25\times 0.1^2}-\frac{0.5}{0.1^2}\right)$

$=4.62\times {10}^{13}\mathrm{N}/\mathrm{C}$