Question

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. Show that the electric field inside this atom is : Ein=Ze4πϵ0(1r^2−rR^3). b. What is the electric field 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 at r = R/2?

Answer 1

Answer:

Part a)

$E = \frac{Ze}{4\pi\epsilon_0}(\frac{1}{r^2} - \frac{r}{R^3})$

Part b)

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

$E = 4.64 \times 10^{13} N/C$

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

$E = \frac{kq}{r^2}$

$E_1 = \frac{Ze}{4\pi \epsilon_0 r^2}$

now charge due to electrons inside a radius "r" is given as

$q = \frac{-Ze r^3}{R^3}$

now we will have electric field given as

$E_2 = \frac{(-Zer^3}{R^3})}{4\pi\epsilon_0 r^2}$

now net electric field is given as

$E = E_1 + E_2$

$E = \frac{Ze}{4\pi \epsilon_0 r^2} - \frac{Zer}{4\pi \epsilon_0 R^3}$

$E = \frac{Ze}{4\pi\epsilon_0}(\frac{1}{r^2} - \frac{r}{R^3})$

Part b)

At the surface of an atom

$r = R$

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

$r = \frac{R}{2}$

so we will have

$E = 92(1.6 \times 10^{-19}) \times (9 \times 10^9)(\frac{4}{R^2} - \frac{1}{2R^2})$

$E = \frac{4.64 \times 10^{-7}}{(0.10 \times 10^{-9})^2}$

$E = 4.64 \times 10^{13} N/C$