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 Fe Fg between an electron and a proton separated by a distance of 3 m. What is the magnitude of the electrical force? The Coulomb constant is 8.98755 × 109 N · m 2 /C 2 , the gravitational constant is 6.67259 × 10−11 m3 /kg · s 2 , the mass of a proton is 1.67262 × 10−27 kg, the mass of an electron is 9.10939 × 10−31 kg, and the elemental charge is 1.602 × 10−19 C. Answer in units of N. What is the ratio of the magnitude of the electrical force to the magnitude of the gravitational force? Answer in units of N.

Answer 1

Answer:

$2.27\cdot 10^{49}$

Explanation:

The gravitational force between the proton and the electron is given by

$F_G=G\frac{m_p m_e}{r^2}$

where

G is the gravitational constant

$m_p$ is the proton mass

$m_e$ is the electron mass

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_G=(6.67259\cdot 10^{-11} m^3 kg s^{-2})\frac{(1.67262\cdot 10^{-27}kg) (9.10939\cdot 10^{-31}kg)}{(3 m)^2}=1.13\cdot 10^{-68}N$

The electrical force between the proton and the electron is given by

$F_E=k\frac{q_p q_e}{r^2}$

where

k is the Coulomb constant

$q_p = q_e = q$ is the elementary charge (charge of the proton and of the electron)

r = 3 m is the distance between the proton and the electron

Substituting numbers into the equation,

$F_E=(8.98755\cdot 10^9 Nm^2 C^{-2})\frac{(1.602\cdot 10^{-19}C)^2}{(3 m)^2}=2.56\cdot 10^{-19}N$

So, the ratio of the electrical force to the gravitational force is

$\frac{F_E}{F_G}=\frac{2.56\cdot 10^{-19} N}{1.13\cdot 10^{-68}N}=2.27\cdot 10^{49}$

So, we see that the electrical force is much larger than the gravitational force.