Question

A hydrogen atom contains a proton, mass 1.67 x 10-27 kilograms, and an electron, mass 9.11 x 10-31 kilograms. The average distance between them is 4.67 x 10-11 meters. The proton and the electron exert both a gravitational force and an electrostatic force on each other.What is the ratio of the greater force to the lesser force?

Answer 1

Answer:

The ratio of the greater force to the lesser force is:

Explanation:

The masses of protons and electron, and the distance between them are given below.

$\begin{gathered} m_p=1.67\times10^{-11}kg \\ \\ m_e=9.11\times10^{-31}kg \\ \\ r=4.67\times10^{-11}kg \\ \\ G=6.67\times10^{-11}m^3kg^{-1}s^{-2} \end{gathered}$

The gravitational force between the proton and electron is calculated below

$\begin{gathered} F_g=\frac{Gm_1m_2}{r^2} \\ \\ F_g=\frac{6.67\times10^{-11}\times1.67\times10^{-27}\times9.11\times10^{-31}}{(4.67\times10^{-11})^2} \\ \\ F_g=4.65\times10^{-47}N \end{gathered}$

The magnitude of the charge on a proton and an electron is the same, and is given as:

$\begin{gathered} q_p=1.6\times10^{-19}C \\ \\ q_e=1.6\times10^{-19}C \end{gathered}$

The electrostatic force between the proton and electron is then calculated as:

$\begin{gathered} F_E=\frac{kq_pq_e}{r^2} \\ \\ F_E=\frac{9\times10^9\times1.6\times10^{-19}\times1.6\times10^{-19}}{(4.67\times10^{-11})^2} \\ \\ F_E=1.056\times10^{-7}N \end{gathered}$

As seen above, the electrostatic force is the greater force, while the gravitational force is the lesser force

The ratio of the greater to the lesser force is:

$\begin{gathered} \frac{F_E}{F_g}=\frac{1.056\times10^{-7}}{4.65\times10^{-47}} \\ \\ \frac{F_E}{F_g}=2.27\times10^{39} \end{gathered}$