Question

The "size" of the atom in Rutherford's model is about 1.0 x 10^−10 m. (a) Determine the attractive electrostatic force between an electron and a proton separated by this distance.
(b) Determine (in eV) the electrostatic potential energy of the atom. (Assume the electron and proton are again separated by the distance stated above.)

Answer 1

Answer:

(a) $2.31\times10^{-8}\ N$

(b) $1.44\times 10^{-19}\ eV$

Explanation:

Given:

*p = charge on proton = $1.602\times 10^{-19}\ C$

*e = magnitude of charge on an electron = $1.602\times 10^{-19}\ C$

*r = distance between the proton and the electron in the Rutherford's atom = $1.0\times 10^{-10}\ m$

Part (a):

Since two unlike charges attract each other.

According to Coulomb's law:

$F=\dfrac{kqe}{r^2}\\\Rightarrow F = \dfrac{9.0\times 10^{9}\times 1.602\times 10^{-19}\times 1.602\times 10^{-19}}{(1.0\times 10^{-10})^2}\\\Rightarrow F = 2.31\times 10^{-8}\ N$

Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is  $2.31\times 10^{-8}\ N$.

Part (b):

Potential energy between two charges separated by a distance r is given by:

$U= \dfrac{kqQ}{r}$

So, the potential energy between the electron and the proton of the Rutherford's atom is given by:

$U = \dfrac{kqe}{r}\\\Rightarrow U = \dfrac{9\times 10^{9}\times1.602\times 10^{-19}\times e}{1.0\times 10^{-10}}\\\Rightarrow U = 1.44\times 10^{-19}\ eV$

Hence, the electrostatic potential energy of the atom is  $1.44\times 10^{-19}\ eV$.