Find the force of attraction between a proton and an electron separated by a distance equal to the radius of the smallest orbit followed by an electron (5 x 10^-11 m) in a hydrogen atom

Respuesta :

Answer:

The answer is [tex] -9.239 \times 10^{-8}\ N = [/tex]

Explanation:

The definition of electric force between two puntual charges is

[tex]F_e = \frac{K q_1 q_2}{d^2}[/tex]

where

[tex]K = 9 \times 10^9\ Nm^2/C^2[/tex].

In this case,

[tex]q_1 = e = 1.602\times 10^{-19}\ C[/tex],

[tex]q_2 = -e = -1.602\times 10^{-19}\ C[/tex]

and

[tex]d = 5 \times 10^{-11}\ m[/tex].

So the force is

[tex]F_e = -9.239 \times 10^{-8}\ N [/tex]

where the negative sign implies force of attraction.