Question

Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of the ratio of the electrical force and gravitational force between an electron and a proton separated by a distance of 1 m.
1. What is the magnitude of the electrical force?
The Coulomb constant is 8.98755 x 10^9 N*m^2/C^2 , the gravitational constant is 6.67259 x 10^11 m^3 /kg*s^2 , the mass of a proton is 1.67262 x 10^-27 kg, the mass of an electron is 9.10939 x 10^−31 kg, and the elemental charge is 1.602 x 10^-19 C. Answer in units of N.

Answer 1

Explanation:

The electrical force between charges is given by :

$F_e=\dfrac{kq_eq_p}{r^2}$

$q_e\ and\ q_p$ are charge on electron and proton respectively.

$F_e=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\\F_e=2.3\times 10^{-28}\ N$

The Gravitational force between masses is given by :

$F_G=\dfrac{Gm_em_p}{r^2}$

$m_e\ and\ m_p$ are masses of electron and proton respectively.

$F_G=\dfrac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}{1^2}\\\\F_G=1.01\times 10^{-67}$

Ratio of electrical to the gravitational force is :

$\dfrac{F_e}{F_G}=\dfrac{2.3\times 10^{-28}\ N}{1.01\times 10^{-67}\ N}\\\\\dfrac{F_e}{F_G}=2.27\times 10^{39}$

Hence, this is the required solution.