Respuesta :
Answer:
Part a)
[tex]F_{attraction} = 1.06 \times 10^{-8} N[/tex]
Part b)
[tex]F_{repulsion} = 1.06 \times 10^{-8} N[/tex]
Explanation:
Part a)
Equilibrium distance is the distance between two centers when two ions just touch each other
So here we will have
[tex]r = 0.068 nm + 0.140 nm[/tex]
[tex]r = 0.208 nm[/tex]
Now Force of attraction between two ions is given by Coulomb's law
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]F = \frac{(9\times 10^9)(1.6 \times 10^{-19})(2\times 1.6 \times 10^{-19})}{(0.208\times 10^{-9})^2}[/tex]
[tex]F_{attraction} = 1.06 \times 10^{-8} N[/tex]
Part b)
At equilibrium separation net force between two ions must be zero
[tex]F_{attraction} = F_{repulsion}[/tex]
so the attraction force and repulsion force must be of equal magnitude
so we have
[tex]F_{repulsion} = 1.06 \times 10^{-8} N[/tex]
