The magnitude of the electrostatic force between two identical ions that are separated by a distance of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathit{m}$is$\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathit{N}$. (a) What is the charge of each ion? (b) How many electrons are “missing” from each ion (thus giving the ion its charge imbalance)?

a) Separation distance of the ions$r=5.0\times {10}^{-10}m$ ,

b) The magnitude of force between the ions $F=3.7x{10}^{-9}N$,

c) They are identical, that is,havingthe same charge,$q$ .

Using the concept of Coulomb's law, we can get the charge on each ion with the given data. As we know, a charged particle emits energy in the form of packets; thus, the number of electrons emitted or present can be considered as a multiple factor of the electronic charge.

Formula:

The magnitude of the electrostatic force between any two particles is$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$. (i)

The number of electrons present is. $n=\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$e$}\right.$(ii)

Let the charge of each ion be $q$. Now, using the magnitude of the force between the ions in equation (i), the charge on each ion is given as(for $q_1=q_2=q$)

localid="1662991888956" width="279">q=rFk=5.0×10-10m3.7×10-9N9.0×109Nm2C2=3.2×10-19C

Hence, the value of the charge on each ion is $3.2\times {10}^{-19}C$.

Let $n$be the number of electrons missing from each ion. Then using the above value of charge on each ion in equation (ii), we can get the value of missing electrons on each ion as follows:

$$\begin{gathered} n=\frac{3.2\times {10}^{-19}C}{1.6\times {10}^{-19}C} \\ =2 \end{gathered}$$

Hence, the value of missing electrons on each ion is $2$.