Question

Calculate the potential energy of interaction between Na and Cl separated by 1 nm in vacuum.

a) 5.69 x 10⁻¹⁸ J
b) 2.31 x 10⁻¹⁸ J
c) 4.63 x 10⁻¹⁸ J
d) 9.26 x 10⁻¹⁸ J

Answer 1

The potential energy of interaction between Na and Cl separated by 1 nm is − 2.31 x 10−19 J, making choice (b) correct.

Step-by-step explanation:

To calculate the potential energy of interaction between a sodium ion (Na+) and a chloride ion (Cl−) separated by a distance of 1 nm in vacuum, we use Coulomb's Law. It is stated as U = (k * q1 * q2) / r, where U is the potential energy, k is Coulomb's constant (8.987 x 109 N m2/C2), q1 and q2 are the charges of the ions, and r is the separation distance in meters. For Na+ and Cl−, q1 = q2 = e, the elementary charge, which is approximately 1.602 x 10−19 C.

Plugging these values into Coulomb's Law, we get U = (8.987 x 109 * (1.602 x 10−19)2) / (1 x 10−9) = 2.31 x 10−19 J. However, since opposite charges attract, the electrostatic potential energy will be negative, U = − 2.31 x 10−19 J. Therefore, the correct choice is (b) 2.31 x 10−19 J.