Question

Calculate the force of attraction between a Ca^2+ and an 0^2- irons whose centers are sep by a distance of 1.25 nm.

Answer 1

Solution:

Given:

Calcium and Oxygen ions each having valence = 2 or 2 valence electrons

seperation distance, r = 1.25nm = $1.25\times 10^{-9}$ m

Coulombian force for two point charges seperated by a distance 'r' is given by:

F = $\frac{1}{4\pi\epsilon _{_{o}} }\times \frac{q_{1}q_{2}}{r^{2}}$      (1)

Now, we know that

$\frac{1}{4\pi \epsilon _{o}} = 9\times 10^{^{9}}$

q = ne

where, n = no. of electrons

e = charge of an electron = $1.6\times 10^{-19}$ C

$q_{1}$ = $n_{1}$e = 2e

$q_{2}$ = $n_{2}$e = 2e

Using Eqn (1)

F = $(9\times 10^{^{9}})$$\times \frac{4\times (1.6\times 10^{-19})^{^{2}}}{(1.25\times 10^{-9})^{2}}$

F = $5.89\times10^{-10}$ N

Since, the force is between two opposite charged ions, it is attractive in nature.