Question

What is the Coulomb force between an electron on the outskirts of the atom and a proton in the nucleus if they are 1.40 x 10^-10m apart?

You are going to calculate the various forces at play in a particular atom. Remember from Chemistry that an atom consists of electrons orbiting a nucleus made up of protons and neutrons.

Answer 1

From Coulomb’s Law force between two charges `q_1and q_2` located at a distance `r` from each other is

`vecF_e = k (q_1*q_2)/r^2hatr`
where k is Coloumb’s constant `=8.99xx10^9\ Nm^2C^-2`

We know that an electron carries the elementary charge `q = -e` and the proton carries the same charge but is positive `q = +e`
`e= 1.602xx10^-19\ C`. Inserting given values in the expression we get

`vecF_e = 8.99xx10^9xx (1.602xx10^-19*(-1.602xx10^-19))/(1.40 xx 10^-10)^2hatr`
`=>vecF_e = -1.06xx10^-5hatr\ N`

negative sign shows that the force is attractive.