Question

The atomic radii of a divalent cation and a monovalent anion are 0.62 nm and 0.121 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).

Answer 1

Answer:

$\boxed{8.4\times 10^{-10} \text{ N}}$

Explanation:

The formula for the force exerted between two ions is  

$F = \dfrac{kq_{1}q_{2}}{r^{2}} = \dfrac{kz_{1}z_{2}e^{2}}{r^{2}}$

where the Coulomb constant

k = 8.988 × 10⁹ N·m·C⁻²

Data:

z₁ = +2; r₁ = 0.62    nm

z₂ =  -1;  r₂ = 0.121 nm

Calculations:

r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm

$F =\dfrac{8.988 \times 10^{9} \text{ N \cdot m ^{2} \cdot C ^{-2} } \times 2 \times (-1) \times ( 1.602 \times 10^{-19} \text{ C})^{2}}{(0.741 \times 10^{-9} \text{ m})^{2}}\\\\ = -\mathbf{8.4\times 10^{-10}} \textbf{ N}\\\text{The attractive force between the ions is } \boxed{\mathbf{8.4\times 10^{-10}} \textbf{ N}}$