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 1

Hint: First write the equation of radius of Bohr orbit ${r_1} = \dfrac{{4\pi {\varepsilon _0}{{\left( {\dfrac{h}{{2\pi }}} \right)}^2}}}{{{m_e}{e^2}}}$ . The electrostatic force of attraction is ${F_C} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}$ and the gravitational force of attraction is ${F_G} = \dfrac{{G{m_p}{m_c}}}{{{r^2}}}$ . Comparing these equations we get $\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}} = G{m_p}{m_c}$ . Replacing this value in the equation of radius of Bohr orbit, we get $r = \dfrac{{{{\left( {\dfrac{h}{{2\pi }}} \right)}^2}}}{{G{m_p}{m_e}}} \approx 1.21 \times {10^{29}}m$ which is much bigger than the known size of the observable universe, i.e. $1.5 \times {10^{27}}m$ .

Complete step by step answer:
Bohr’s model of a hydrogen atom depicts the structure of an atom in reference to a hydrogen atom. According to Bohr’s model of a hydrogen atom in the center of an atom lies a nucleus that contains all the nucleons (the protons and the neutrons). The electrons revolve around the nucleus in fixed circular stationary objects, in these stationary orbits, the electrons have a fixed speed and hence angular momentum.
We know that the radius of the first Bohr orbit is given by the relation,
${r_1} = \dfrac{{4\pi {\varepsilon _0}{{\left( {\dfrac{h}{{2\pi }}} \right)}^2}}}{{{m_e}{e^2}}}$ (Equation 1)
In this equation, ${r_1} = $ The radius of the first Bohr orbit
${\varepsilon _0} = $ The permittivity of free space
$h = $ Planck’s constant
${m_e} = $ The mass of an electron
$e = $ The charge of an electron
${m_p} = $ The mass of a proton
The electrostatic force of attraction between an electron and a proton is given as:
${F_C} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}}$
The gravitational force of attraction between an electron and a proton is given as;
${F_G} = \dfrac{{G{m_p}{m_c}}}{{{r^2}}}$
If the electrostatic force of attraction and the gravitational force of attraction are equal then
${F_C} = {F_G}$
$\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r^2}}} = \dfrac{{G{m_p}{m_c}}}{{{r^2}}}$
Here, $G = 6.67 \times {10^{11}}N{m^2}/k{g^2} = $ Gravitational constant
$\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}} = G{m_p}{m_c}$
Substitute the value of $\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}$ in equation 1, we get
$r = \dfrac{{{{\left( {\dfrac{h}{{2\pi }}} \right)}^2}}}{{G{m_p}{m_e}}}$
$r = \dfrac{{{{\left( {\dfrac{{6.63 \times {{10}^{ - 34}}}}{{2 \times 3.14}}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}} \times 1.67 \times {{10}^{ - 27}} \times {{\left( {9.1 \times {{10}^{ - 31}}} \right)}^2}}}$
$r \approx 1.21 \times {10^{29}}m$
We know that the universe is 156 billion years (one light-year is the distance that light covers in one year) or $1.5 \times {10^{27}}m$ wide.
So the radius of the first Bohr orbit is bigger than the estimated size of the whole universe.

Note: The result that we reached in the solution suggests that no gravitational force works on the sub-atomic level, because the radius of first Bohr first orbit cannot be greater than the radius of the observable universe, this means that only electrostatic force of attraction and nuclear forces work on the sub-atomic level.