Question

g The “size” of the atom in Rutherford’s model is about 8 × 10−11 m. Determine the attractive electrostatics force between a electron and a proton separated by this distance. Answer in units of N.

Answer 1

Answer:

$3.6\cdot 10^{-8} N$

Explanation:

The electrostatic force between the proton and the electron is given by:

$F=k\frac{q_p q_e}{r^2}$

where

$k=9.00\cdot 10^9 Nm^2 C^{-2}$ is the Coulomb constant

$q_p = 1.6\cdot 10^{-19} C$ is the magnitude of the charge of the proton

$q_e = 1.6\cdot 10^{-19}C$ is the magnitude of the charge of the electron

$r=8\cdot 10^{-11}m$ is the distance between the proton and the electrons

Substituting the values into the formula, we find

$F=(9\cdot 10^9 ) \frac{(1.6\cdot 10^{-19})^2}{(8\cdot 10^{-11})^2}=3.6\cdot 10^{-8} N$