Question

Calculate the force of attraction between a K and an O2 ion the centers of which are separated by a distance of 1.5 nm.

Answer 1

Answer:

Force of attraction, $F=2.048\times 10^{-10}\ N$

Explanation:

The charge on potassium ion $K^+$, $q_1=1.6\times 10^{-19}\ C$

The charge on the oxygen ion $O^{-2}$, $q_2=2\times 1.6\times 10^{-19}=3.2\times 10^{-19}\ C$

Distance between ions, $r=1.5\ nm=1.5\times 10^{-9}\ m$

To find,

The force of attraction between a  $K^+$ and an  $O^{-2}$ ion.

Solution,

The force of attraction between ions is given by the electric force of attraction. It is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$F=\dfrac{9\times 10^9\times 1.6\times 10^{-19}\times 2\times 1.6\times 10^{-19}}{(1.5\times 10^{-9})^2}$

$F=2.048\times 10^{-10}\ N$

So, the force of attraction between ion is $F=2.048\times 10^{-10}\ N$.