Question

Calculate the force of gravitational pull between the electron and the proton on the hydrogen atom. It is known that the radius of the electron orbit is R = 0.5 × 10 ‐ ¹⁰m, the mass of the electron m1 = 9.1 × 10 ‐ ³¹ kg and the mass of the proton m2 = 1.67 × 10 ‐ ²⁷ kg. Compare the conclusion you will find with the value of the electric gravitational force, which is Fel = 9 × 10 ‐⁸ N. What conclusion can you formulate?

Answer 1

By the Newton's Universal Gravitation Law, we have:

$$F_g=\dfrac{Gm_em_p}{R^2},$$

$F_g=\dfrac{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot9.1\cdot10^{-31}\ kg\cdot1.67\cdot10^{-27}\ kg}{(0.5\cdot10^{-10}\ m)^2}=4.05\cdot10^{-47}\ N.$

Let's calculate the ratio of the electrostatic force and fravitational force between the electron and the proton:

$$\dfrac{F_e}{F_g}=\dfrac{9.0\cdot10^{-8}\ N}{4.05\cdot10^{-47}\ N}=2.2\cdot10^{39}.$$

Therefore, the electric force between electron and proton is $2.2\cdot10^{39}$ times bigger than the gravitational force.