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)  8.2 x 10^-8 N

(b) 3.6 x 10^-47 N  , 2.27 x 10^39

Explanation:

charge of proton, q1 = 1.6 x 10^-19 C

charge of electron, q2 = - 1.6 x 10^-19 C

radius of orbit, r = 0.053 nm = 0.053 x 10^-9 m

mass of electron, me = 9.1 x 10^-31 kg

mass of proton, mp = 1.67 x 10^-27 kg

Gravitational constant, G = 6.67 x 10^-11 Nm^2/kg^2

(a) The electrostatic force between two charges is given by

Where, K is the coulombic constant = 9 x 10^9 Nm^2/C^2

By substituting the values

Fe = 8.2 x 10^-8 N

(b) The gravitational force between the electron and proton is given by

Fg = 3.6 x 10^-47 N

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this