Question

In the Bohr model of the atom, atomic electrons approximatately 'orbit' the nucleus. The hydrogen atom consists of a proton of mass 1.67 × 10-27 kg and an orbiting electron of mass 9.11 × 10-31 kg. In one of its orbits, the electron is 5.3 × 10-11 m from the proton. What is the mutual attractive gravitational force between the electron and proton

Answer 1

Answer:

$F=3.61\times 10^{-47}\ N$

Explanation:

Mass of a proton, $m_p=1.67\times 10^{-27}\ kg$

Mass of an electron, $m_e=9.11\times 10^{-31}\ kg$

The distance between the electron and the proton is, $r=5.3\times 10^{-11}\ m$

We need to find the mutual attractive gravitational force between the electron and proton. The gravitational force is given by :

$F=G\dfrac{m_em_p}{r^2}$

Where G is the universal Gravitational constant

$F=6.67\times 10^{-11}\times \dfrac{9.11\times 10^{-31}\times 1.67\times 10^{-27}}{(5.3\times 10^{-11})^2}\\\\F=3.61\times 10^{-47}\ N$

So, the force between the electron and proton is $3.61\times 10^{-47}\ N$.