Question

The electrostatic force of repulsion between two equal positively charged ions is $3.7\times {10}^{-9}N$, when they are separated by a distance of $5\stackrel{\circ }{A}$. How many electrons are missing from each ion?

Answer 1

Given, the force of repulsion between two equally positively charged ions $=3.7\times {10}^{-19}N$

Distance of separation, $d=5$Å $=5\times {10}^{10}m$

To find the number of electrons missing from each ion.

Let us suppose charge is $q_1=q_2=q$

Let $n$ be the number of electrons missing.

Using Coulomb's Law, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{d^2}$

$F=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{d^2}$

So, $q^2=4\pi {\epsilon }_0\times F\times d^2$

$=\frac{1}{9\times {10}^9}\times 3.7\times {10}^{-9}\times (5\times {10}^{-10}{)}^2$

$=10.28\times {10}^{-38}$

$q=3.2\times {10}^{-19}C$

As, $q=ne$

So, $n=\frac{q}{e}=\frac{3.2\times {10}^{-19}}{1.6\times {10}^{-19}}$

$n=2$