Question
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Answer

Radius of the first Bohr orbit is given by the relation,
$\text{r}_1=\frac{4\pi\in_0\Big(\frac{\text{h}}{2\pi}\Big)^2}{\text{m}_\text{e}\text{e}^2}\ \dots(1)$
Where,
$\in_0$ = Permittivity of free space
h = Planck ' s constant = 6.63 × 10-34 Js
me = Mass of an electron = 9.1 × 10-31 kg
e = Charge of an electron = 1.9 × 10-19 C
mp = Mass of a proton = 1.67 × 10-27 kg
r = Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
$\text{F}_\text{C}=\frac{\text{e}^2}{4\pi\in_0\text{r}^2}\ \dots(2)$
Gravitational force of attraction between an electron and a proton is given as:
$\text{F}_\text{G}=\frac{\text{Gm}_\text{p}\text{m}_\text{c}}{\text{r}^2}\ \dots(3)$
Where,
G = Gravitational constant = 6.67 × 10-11 N m/ kg2
If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:
$\therefore\ \text{FG}=\text{FC}$
$\frac{\text{Gm}_\text{p}\text{m}_\text{c}}{\text{r}^2}=\frac{\text{e}^2}{4\pi\in_0\text{r}^2}$
$\therefore\ \frac{\text{e}^2}{4\pi\in_0}=\text{Gm}_\text{p}\text{m}_\text{c}\ \dots(4)$
Putting the value of equation (4) in equation (1), we get:
$\text{r}_1=\frac{\Big(\frac{\text{h}}{2\pi}\Big)^2}{\text{Gm}_\text{p}\text{m}_\text{e}^2}$
$=\frac{\Big(\frac{6.63\times10^{-34}}{2\times3.14}\Big)^2}{6.67\times10^{-11}\times1.67\times10^{-27}\times(9.1\times10^{-31})^2}\approx1.21\times10^{29}\text{ m}$
It is known that the universe is 156 billion light years wide or 1.5 × 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

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