Chapter 20, Problem 56A

To determine

Electrostatic force between an electron and a proton.

Answer to Problem 56A

  $F=8.2\times {10}^{-8}\text{N}$ attractive

Explanation of Solution

Given:

An electron and a proton in hydrogen atom.

Formula Used:

From Coulomb’s law,

  $F=\frac{K{q}_1{q}_2}{d^2}$

Where,

  $F$ is electrostatic force

  $K$ is coulomb’s constant

  $q_1$ and $q_2$ are charges and

  $d$ is distance between the charges.

Calculation:

The charge on electron is, $q_1=-1.6\times {10}^{-19}\text{C}$

The charge on proton is, $q_2=1.6\times {10}^{-19}\text{C}$

Approximate radius of hydrogen atom is, $d=53\text{pF}$

As, $K=9\times {10}^9\text{N}{\text{.m}}^2/{\text{C}}^2$

On putting the values, the force is

  $\begin{array}{l}F=\frac{9\times {10}^9\times \left(-1.6\times {10}^{-19}\right)\times \left(1.6\times {10}^{-19}\right)}{{\left(53\times {10}^{-12}\right)}^2}\\ F=-8.2\times {10}^{-8}\text{N}\end{array}$

Negative sign represents that the force is attractive in nature.

Conclusion:

Therefore, the electrostatic force between an electron and a proton in hydrogen atom is

  $F=8.2\times {10}^{-8}\text{N}$ attractive.