The electron orbiting the proton in the hydrogen atom is at a distance of $5.3 \times 10^{-9} \mathrm{~m}$. What is the ratio of the electric force to the gravitational force between these particles? (The mass of the proton is $1.67 \times 10^{-27} \mathrm{~kg}$ and the mass of the electron is $9.11 \times 10^{-31} \mathrm{~kg}$.)

Question

Quizplus · Accepted Answer

The electric force $ F_e $ is given by Coulomb's law $ F_e = \frac{k \cdot e^2}{r^2} $, and the gravitational force $ F_g $ is given by Newton's law of gravitation $ F_g = \frac{G \cdot m_p \cdot m_e}{r^2} $. The ratio $ \frac{F_e}{F_g} = \frac{k \cdot e^2}{G \cdot m_p \cdot m_e} $ is approximately $ 2.3 \times 10^{39} $.

The Electron Orbiting the Proton in the Hydrogen Atom Is 5.3×10−9 m5.3 \times 10^{-9} \mathrm{~m}5.3×10−9 m

The Electron Orbiting the Proton in the Hydrogen Atom Is $5.3 \times 10^{-9} \mathrm{~m}$