Answer:
[tex]\large \boxed{\text{(a) }-1.10 \times 10^{-8} \text{ N}; \text{(b) }+1.10 \times 10^{-8} \text{ N}}[/tex]
Explanation:
(a) Attractive force
To calculate the attractive force (F), we can use Coulomb's Law:
[tex]F =k\dfrac{Q_{1}Q_{2}}{r^{2}}[/tex]
where
Q₁ and Q₂ are the charges on the ions,
r is the distance between them, and
k = the Coulomb constant
Data:
Q₁ = 2+
Q₂ = 1-
r₊ = 0.072 nm
r₋ = 0.133 nm
k = 8.988 × 10⁹ N· m²C⁻²
Calculations:
Q₁ = (+2) × 1.602 × 10⁻¹⁹ C = +3.204 × 10⁻¹⁹ C
Q₂ = (-1) × 1.602 × 10⁻¹⁹ C = -1.602 × 10⁻¹⁹ C
r = r₊ + r₋ = 0.072 nm + 0.133 nm = 0.205 nm = 0.205 × 10⁻⁹ m
[tex]\begin{array}{rcl}F & = & 8.988 \times 10^{9}\times\dfrac{3.204 \times 10^{-19} \times (-1.602) \times 10^{-19}}{(0.205 \times 10^{-9})^{2}}\\\\& = & \mathbf{-1.10 \times 10^{-8}} \textbf{ N}\\\end{array}\\\text{The attractive force between the two ions is $\large \boxed{\mathbf{-1.10 \times 10^{-8}} \textbf{ N}}$}[/tex]
(b) Repulsive force
The equilibrium position is reached when the ions just touch each other. If they come any closer, the nuclear repulsions will outweigh the Coulomb attraction.
Thus, the repulsive force is equal and opposite to the attractive force.
[tex]\text{The repulsive force is $\large \boxed{\mathbf{+1.10 \times 10^{-8}} \textbf{ N}}$}[/tex]
