Answer to Question #167382 in Physics for Isey

Question

Calculate the force of attraction between the lone electron and proton in a hydrogen atom. The average distance between them is 5.0 x 10 ^ -11 *m.

Answer 1

We can find the force of attraction from the Coulomb's law:

$$F=\dfrac{kq_1q_2}{r^2},$$ $$F=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(-1.6\cdot10^{-19}\ C)\cdot1.6\cdot10^{-19}\ C}{(5\cdot10^{-11}\ m)^2}=-9.22\cdot10^{-8}\ N.$$

The magnitude of the electrostatic force between the electron and proton is $9.22\cdot10^{-8}\ N$. Since the two charges are opposite in sign they will attract each other, therefore, the direction of the electrostatic force is attractive, hence, the force between electron and proton is attractive (also, the sign minus indicates that the direction and force are attractive).