Question

In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single electron orbits around the nucleus. What is the ratio of the electric force and the gravitational force acting between the proton and the electron, when they are 4.53 angstrom away from each other? (Possibly useful constants: Coulomb constant: k = 8.9876×109 N*m2/C2, Universal Gravitational constant: G = 6.6726×10-11 N*m2/kg2, elementary charge: e = 1.6022×10-19 C, electron mass: me = 9.1094×10-31 kg, proton mass: mp = 1.6726×10-27 kg. You might also find the specific charges i.e. charge per mass ratios of the electron and the proton useful: electron: e/me = 1.7588×1011 C/kg; proton: e/mp = 9.5788×107 C/kg.)

Answer 1

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$. Hence, this is the required solution.