Question

The atomic radii of Li+and O2−ions are 0.068 and 0.140 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).(2 points)(b)What is the force of repulsion at this same separation distance?(1 point)

Answer 1

Answer:

Part a)

$F_{attraction} = 1.06 \times 10^{-8} N$

Part b)

$F_{repulsion} = 1.06 \times 10^{-8} N$

Explanation:

Part a)

Equilibrium distance is the distance between two centers when two ions just touch each other

So here we will have

$r = 0.068 nm + 0.140 nm$

$r = 0.208 nm$

Now Force of attraction between two ions is given by Coulomb's law

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

$F = \frac{(9\times 10^9)(1.6 \times 10^{-19})(2\times 1.6 \times 10^{-19})}{(0.208\times 10^{-9})^2}$

$F_{attraction} = 1.06 \times 10^{-8} N$

Part b)

At equilibrium separation net force between two ions must be zero

$F_{attraction} = F_{repulsion}$

so the attraction force and repulsion force must be of equal magnitude

so we have

$F_{repulsion} = 1.06 \times 10^{-8} N$