Question

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

Answer 1

Answer:

$F=2.04\times 10^{-10} N$

Explanation:

Given that

$q_1=1\times 1.6\times 10^{-19}$

$q_2=2\times 1.6\times 10^{-19}$

r= 1.5 nm

We know that force between two charge given as follows

$F=K\dfrac{q_1q_2}{r^2}$

Where K is the constant

Noe by putting the values

$q_1\times q_2=2\times 1.6\times1.6\times 10^{-38}$

$F=8.98\times 10^9 \times \dfrac{5.12\times 10^{-38}}{\left ( 1.5\times10^{-9} \right )^2}$

$F=2.04\times 10^{-10} N$