Chapter 20, Problem 34A

To determine

Electrostatic force between two electrons in an atom.

Answer to Problem 34A

  $F=1.0\times {10}^{-8}\text{N}$ repulsive in nature.

Explanation of Solution

Given:

Two electrons in an atom.

Distance between the electrons, $d=1.5\times {10}^{-10}\text{m}$

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:

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

And charge on electrons is $1.6\times {10}^{-19}\text{C}$

So, $q_1=q_2=-1.6\times {10}^{-19}\text{C}$

On putting the values,

  $\begin{array}{c}F=\frac{9\times {10}^9\times \left(-1.6\times {10}^{-19}\right)\times \left(-1.6\times {10}^{--19}\right)}{{\left(1.5\times {10}^{-10}\right)}^2}\\ =1.0\times {10}^{-8}\text{N}\end{array}$

As, $F=1.0\times {10}^{-8}\text{N}$ it means the force is repulsive in nature that is away from each other.

Conclusion:

Therefore, the electrostatic force between the electrons in atom is $F=1.0\times {10}^{-8}\text{N}$ away from each other.