Question

Calculate the electrostatic force of attraction between a proton and an electron in a hydrogen atom. The radius of the electron orbit is $0.05{\text{ nm}}$ and charge on the electron is $1.6 \times {10^{ - 19}}C$ .

Answer 1

In a hydrogen atom, there is only one proton inside the nucleus and one electron which revolves around the nucleus in an orbit. This electron will always be at a fixed distance (equal to the radius of the orbit) from the proton.

Formula Used:
The electrostatic force between any two charges ${q_1}$ and ${q_2}$ separated by a distance $r$ is given by $F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$ where $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$ which has value $9 \times {10^9}{\text{ N}}{{\text{C}}^{ - 2}}{{\text{m}}^2}$ .

Complete step by step answer:
Let us first discuss a hydrogen atom. In a hydrogen atom, there is only one proton inside the nucleus and one electron which revolves around the nucleus in an orbit. This electron will always be at a fixed distance (equal to the radius of the orbit) from the proton.
We know that the electrostatic force between any two charges ${q_1}$ and ${q_2}$ separated by a distance $r$ is given by $F = \dfrac{{K{q_1}{q_2}}}{{{r^2}}}$ where $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$ which has value $9 \times {10^9}{\text{ N}}{{\text{C}}^{ - 2}}{{\text{m}}^2}$ .
As from the question, the two charges are the charges on an electron and a proton.
The magnitude of charge on electron and proton is same and equal to $1.6 \times {10^{ - 19}}C$ i.e. ${q_1} = {q_2} = 1.6 \times {10^{ - 19}}C$. The distance between the two charges is $r = 0.05{\text{ nm}} = 0.05 \times {10^{ - 9}}m$ .
So, the electrostatic force between the two charges is given by
$F = \dfrac{{9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{{{\left( {0.05 \times {{10}^{ - 9}}} \right)}^2}}}$
On further solving we have
$F = \dfrac{{9 \times 1.6 \times 1.6 \times {{10}^{ - 29}}}}{{25 \times {{10}^{ - 22}}}}$
On simplifying we have
$F = 9.216 \times {10^{ - 8}}{\text{ N}}$
Hence, the electrostatic force of attraction between a proton and an electron in a hydrogen atom is $9.216 \times {10^{ - 8}}{\text{ N}}$ .

Note: Although the magnitude of the charge on an electron and a proton is the same but their signs are opposite of each other. The charge on a proton is positive whereas the charge on an electron is negative. As the polarity of both the charges is opposite, that’s why the electrostatic force between them is attractive in nature. Always remember like charges repel each other and unlike charges attract each other.