In the most common isotope of Hydrogen the nucleus is made out of a single proto

Question

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single electron orbits around the nucleus. What is the ratio of the electric force and the gravitational force acting between the proton and the electron, when they are 2.71 angstrom away from each other? (Possibly useful constants: Coulomb constant: k = 8.9876×109 N*m2/C2, Universal Gravitational constant: G = 6.6726×10-11 N*m2/kg2, elementary charge: e = 1.6022×10-19 C, electron mass: me = 9.1094×10-31 kg, proton mass: mp = 1.6726×10-27 kg. You might also find the specific charges i.e. charge per mass ratios of the electron and the proton useful: electron: e/me = 1.7588×1011 C/kg; proton: e/mp = 9.5788×107 C/kg.)

Explanation / Answer

Electric force, Fe = kqprotonqelectron / r2

where qproton and qelectron are charges of proton and electron, respectively.

r is distance between these two = 2.71 angstrom

as 1 angstrom = 1e-10 m

so, 2.71 angstrom = 2.71e-10 m

Fe = 8.9876×109 N*m2/C2 * 1.6e-19 C * 1.6e-19 C / (2.71e-10)2

Fe = 3.1328e-9 N

Now, Force of gravitation

Fg = Gmprotonmelectron / r2

Fg = 6.6726×10-11 N*m2/kg2*1.67e27 Kg * 9.1e-31 Kg / (2.71e-10)2

Fg = 1.381e-48 N

ratio of the electric force and the gravitational force

Fe / Fg = 3.1328e-9 N / 1.381e-48 N

Fe / Fg = 2.2685e39

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