Question

An early planetary model of the hydrogen atom consisted of a 1.67 X 10-27 kg proton in the nucleus and a 9.11 X 10-31 kg electron in orbit around it at a distance of 5.0 X 10-11 m. In this model, what is the gravitational force between a proton and an electron?

Answer 1

Answer:

The gravitational force between proton and an electron is $4.06 \times 10^{-47}$ N

Explanation:

Given:

Mass of proton $m_{1} = 1.67 \times 10^{-27}$ kg

Mass of electron $m_{2} = 9.11 \times 10^{-31}$ kg

Separation between electron and proton $r = 5 \times 10^{-11}$ m

According to the gravitational law,

   $F = \frac{Gm_{1} m_{2} }{r^{2} }$

Where $G = 6.674 \times 10^{-11}$ = gravitational constant.

   $F = \frac{6.674 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31} }{(5 \times 10^{-11} )^{2} }$

   $F = 4.06 \times 10^{-47}$ N

Therefore, the gravitational force between proton and an electron is $4.06 \times 10^{-47}$ N