Problem — Coulomb force in the hydrogen atom

Problem

In the planetary model of the hydrogen atom, the electron travels an orbit of radius $r \approx 5{,}3\cdot 10^{-11}\ \text{m}$ around the proton. Calculate the magnitude of the Coulomb force between the particles and compare it with the gravitational force ($m_e = 9{,}1\cdot 10^{-31}\ \text{kg}$, $m_p = 1{,}67\cdot 10^{-27}\ \text{kg}$): by how many orders of magnitude is the electric force greater?

Solution

Electric force. The electron and proton have charge $\pm e$ with $e = 1{,}6\cdot 10^{-19}\ \text{C}$. Coulomb’s law with $k = 8{,}99\cdot 10^9\ \text{N·m}^2/\text{C}^2$:

$F_e = k\,\frac{e^2}{r^2} = 8{,}99\cdot 10^9 \cdot \frac{(1{,}6\cdot 10^{-19})^2}{(5{,}3\cdot 10^{-11})^2} = 8{,}99\cdot 10^9 \cdot \frac{2{,}56\cdot 10^{-38}}{2{,}81\cdot 10^{-21}} \approx 8{,}2\cdot 10^{-8}\ \text{N}$

Gravitational force. With $G = 6{,}67\cdot 10^{-11}\ \text{N·m}^2/\text{kg}^2$:

$F_g = G\,\frac{m_e m_p}{r^2} = 6{,}67\cdot 10^{-11} \cdot \frac{(9{,}1\cdot 10^{-31})(1{,}67\cdot 10^{-27})}{2{,}81\cdot 10^{-21}} = 6{,}67\cdot 10^{-11} \cdot \frac{1{,}52\cdot 10^{-57}}{2{,}81\cdot 10^{-21}} \approx 3{,}6\cdot 10^{-47}\ \text{N}$

Ratio:

$\frac{F_e}{F_g} = \frac{8{,}2\cdot 10^{-8}}{3{,}6\cdot 10^{-47}} \approx 2{,}3\cdot 10^{39}$

The electric force is about $39$ orders of magnitude stronger than the gravitational one: inside the atom, gravity is entirely negligible.

$\ev{F_e\approx 8{,}2\cdot 10^{-8}\ \text{N},\ F_e/F_g\sim 10^{39}}$

Links

Topics: Elettrostatica Concepts: Legge di Coulomb Objects: Atomo di idrogeno