In a hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm. (a) find the magnitude of the electrostatic force of attraction, fe between the electron and the proton. (b) find the magnitude of the gravitational force of attraction fg , between the electron and the proton, and find the ratio, fe /fg . me = 9.11 x 10-31kg, e = 1.602 x 10-19c mp = 1.67 x 10-27kg k = 9 x 109 nm2 /c2 g = 6.67 x 10-11 nm2 /kg2

(a):

(b):

(c):

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges and respectively is given by

where,

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom,

The charge on the electron,

These two are separated by the distance,

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

Part (b):

The gravitational force of attraction between two objects of masses and respectively is given by

where,

For the given system,

The mass of proton,

The mass of the electron,

Distance between the two,

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

The ratio :

the answer would be 32*f