Question

Calculate the energy of attraction between a cation with a valence of 1 and an anion with a valence of 3 the centers of which are separated by a distance of 7.5 nm

Answer 1

Answer:

The energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J

Explanation:

Let the charge on the cation be q₁

Also let the charge on the anion be q₂

A cation q₁ with a valence of 1, has a charge of 1 X 1.602×10⁻¹⁹C = 1.602×10⁻¹⁹C

An anion q₂ with a valence of 3, has a charge of 3 X 1.602×10⁻¹⁹C = 4.806 ×10⁻¹⁹C

The distance between the two charges is 7.5nm = 7.5 X10⁻⁹m

Energy of attraction = $\frac{Kq_1q_2}{r^2}$

Where k is coulomb's constant = 8.99 X 10⁹ Nm₂/C₂

Energy of attraction = $\frac{8.99 X 10^9 (1.602X10^{-19})(4.806 X10^{-19})}{(7.5X10^{-9})^2}$

Energy of attraction = 1.231 X 10⁻¹¹ J

Therefore, the energy of attraction between the cation and anion is 1.231 X 10⁻¹¹ J