The gravitational attraction between electron and proton

Solution:

Radius of the first Bohr orbit is given by the relation,

$r_{1}=\frac{4 \pi \in_{0}\left(\frac{h}{2 \pi}\right)^{2}}{m_{e} e^{2}}$     ...(1)

Where,

∈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: $F_{\mathrm{C}}=\frac{e^{2}}{4 \pi \epsilon_{0} r^{2}}$   ..(2)

Gravitational force of attraction between an electron and a proton is given as:

$F_{\mathrm{G}}=\frac{\mathrm{G} m_{\mathrm{p}} m_{\mathrm{e}}}{r^{2}}$   ..(3)

Where,

G = Gravitational constant = 6.67 × 10−11 N m2/kg2

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

$\therefore F_{G}=F_{C}$

$\frac{\mathrm{G} m_{\mathrm{p}} m_{\mathrm{e}}}{r^{2}}=\frac{e^{2}}{4 \pi \in_{0} r^{2}}$

$\therefore \frac{e^{2}}{4 \pi \epsilon_{0}}=\mathrm{G} m_{\mathrm{p}} m_{\mathrm{e}}$      ...(4)

Putting the value of equation (4) in equation (1), we get:

$r_{1}=\frac{\left(\frac{h}{2 \pi}\right)^{2}}{G m_{p} m_{e}{ }^{2}}$

$=\frac{\left(\frac{6.63 \times 10^{-34}}{2 \times 3.14}\right)^{2}}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times\left(9.1 \times 10^{-31}\right)^{2}} \approx 1.21 \times 10^{29} \mathrm{~m}$

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.