Question

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x10-27 kg). Assume that the radius of the orbit is 2.5 Angstroms (10-10 m). (a) What is the magnitude of the gravitational force of attraction between electron and proton? AN (b) What is the magnitude of the electric force of attraction between electron and proton? N

Answer 1

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$