Respuesta :
Answer:
(a) [tex]2.31\times10^{-8}\ N[/tex]
(b) [tex]1.44\times 10^{-19}\ eV[/tex]
Explanation:
Given:
*p = charge on proton = [tex]1.602\times 10^{-19}\ C[/tex]
*e = magnitude of charge on an electron = [tex]1.602\times 10^{-19}\ C[/tex]
*r = distance between the proton and the electron in the Rutherford's atom = [tex]1.0\times 10^{-10}\ m[/tex]
Part (a):
Since two unlike charges attract each other.
According to Coulomb's law:
[tex]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[/tex]
Hence, the attractive electrostatic force of attraction acting between an electron and a proton of Rutherford's atom is  [tex]2.31\times 10^{-8}\ N[/tex].
Part (b):
Potential energy between two charges separated by a distance r is given by:
[tex]U= \dfrac{kqQ}{r}[/tex]
So, the potential energy between the electron and the proton of the Rutherford's atom is given by:
[tex]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[/tex]
Hence, the electrostatic potential energy of the atom is  [tex]1.44\times 10^{-19}\ eV[/tex].
